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Algebraic Identities

Reading Time: 45 mins

Solved Examples: 20

Practice Questions: 5

What are Algebraic Identities?

Algebraic identities are equations that are true for all permissible values of the variables involved. Unlike ordinary algebraic equations, which are true only for specific values of the variables, identities hold universally because both sides of the equation are always equal. Algebraic identities provide standard formulas for expanding, simplifying and factorising algebraic expressions. Some of the most commonly used algebraic identities are:

(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²
(a + b)(a − b) = a² − b²
(x + a)(x + b) = x² + (a + b)x + ab
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac
(a + b)³ = a³ + b³ + 3ab(a + b)
(a − b)³ = a³ − b³ − 3ab(a − b)
(a + b + c)(a² + b² + c² − ab − bc − ca) = a³ + b³ + c³ − 3abc
(a + b)(a + c)(b + c) = (a + b + c)(ab + ac + bc) − abc
a² + b² + c² = (a + b + c)² − 2(ab + ac + bc)

Competitive Exam Shortcuts

For expressions involving a + 1/a, generate higher powers quickly

If:
a + 1/a = n
then:
a² + 1/a² = n² − 2
a³ + 1/a³ = n³ − 3n
a⁴ + 1/a⁴ = n⁴ − 4n² + 2

For expressions involving a + b and ab, use cube identities immediately

If: a + b = S and ab = P
then:
a³ + b³ = S³ − 3PS
and
a³ − b³ = (a − b)(S² − P)
This is much faster than finding individual values of a and b.

Convert squares into sums and products

When given:
a² + b²
and
a + b
use:
a² + b² = (a + b)² − 2ab
When given:
ab and a + b, immediately think of this identity.

For three variables, remember the key identity

For questions involving a, b, c:
a³ + b³ + c³ − 3abc
directly becomes:
(a + b + c)(a² + b² + c² − ab − bc − ca)

For expressions with square roots, remove the root first

If √a + 1/√a = n
Square first
a + 1/a = n² − 2
Then proceed to higher powers.

Check for hidden common factors

Expressions like
a⁴ + 1/a²
can often be rewritten as
a(a³ + 1/a³)
which connects directly to a known identity.

Summary of Algebraic Identities

Identity	Formula
Square of Sum	(a + b)² = a² + 2ab + b²
Square of Difference	(a − b)² = a² − 2ab + b²
Difference of Squares	(a + b)(a − b) = a² − b²
Product of Two Binomials	(x + a)(x + b) = x² + (a + b)x + ab
Square of Three Terms	(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac
Cube of Sum	(a + b)³ = a³ + b³ + 3ab(a + b)
Cube of Difference	(a − b)³ = a³ − b³ − 3ab(a − b)
Sum of Three Cubes Identity	(a + b + c)(a² + b² + c² − ab − bc − ca) = a³ + b³ + c³ − 3abc
Product of Three Binomials	(a + b)(a + c)(b + c) = (a + b + c)(ab + ac + bc) − abc
Sum of Squares of Three Terms	a² + b² + c² = (a + b + c)² − 2(ab + ac + bc)

Solved Examples

Question 1: If a + b + c = 9 and ab + bc + ca = 26, find the value of a² + b² + c²

a² + b² + c² = (a + b + c)² − 2(ab + bc + ca)
= 9² − 2 × 26
= 81 − 52
= 29

Question 2: If a + b + c = 6, ab + bc + ca = 11 and abc = 6, find the value of a³ + b³ + c³

Using the identity:
a³ + b³ + c³
= (a + b + c)(a² + b² + c² − ab − bc − ca) + 3abc
First,
a² + b² + c²
= (a + b + c)² − 2(ab + bc + ca)
= 6² − 2 × 11
= 36 − 22
= 14
Now,
a³ + b³ + c³
= 6(14 − 11) + 3 × 6
= 18 + 18
= 36

Question 3: If 3x + 1/(3x) = 5 then find the value of: 9x² + 1/(9x²)

Given: 3x + 1/(3x) = 5
Squaring both sides:
(3x + 1/(3x))² = 5²
Using the identity:
(a + b)² = a² + 2ab + b²
(3x)² + 2 × 3x × 1/(3x) + (1/(3x))² = 25
9x² + 2 + 1/(9x²) = 25
Therefore,
9x² + 1/(9x²) = 25 − 2
= 23

Question 4: If x − 1/x = 2 then find the value of x³ − 1/x³

A number is divisible by 12 if it is divisible by 3 and 4.
Divisibility by 4: The last two digits are 24, which is divisible by 4.
Divisibility by 3: The sum of the digits is:
5 + 8 + k + 3 + 2 + 4 = 22 + k
For divisibility by 3,
22 + k must be divisible by 3.
Only k = 2, 5, 8 satisfy this condition.
Answer: k = 2, 5, 8

Question 5: If a − 1/a = 2 then find the value of: a³ − 1/a³

Given: a − 1/a = 2
Cubing both sides:
(a − 1/a)³ = 2³
Using the identity:
(p − q)³ = p³ − q³ − 3pq(p − q)
Here,
p = a and q = 1/a
Therefore,
a³ − 1/a³ − 3 × a × 1/a × (a − 1/a) = 8
Since,
a × 1/a = 1
Therefore,
a³ − 1/a³ − 3 × 1 × 2 = 8
a³ − 1/a³ − 6 = 8
a³ − 1/a³ = 14

Question 6: If √a + 1/√a = 2, then find the value of 5a² + 1/(5a²).

Let: x = √a
Therefore,
x + 1/x = 2
Multiplying both sides by x:
x² + 1 = 2x
x² − 2x + 1 = 0
(x − 1)² = 0
x = 1
Therefore,
√a = 1
a = 1
Now,
5a² + 1/(5a²)
= 5 × 1² + 1/(5 × 1²)
= 5 + 1/5
= 26/5

Question 7: If a³ − b³ = 56 and a − b = 2 then find the value of: a² + b²

Using the identity:
a³ − b³ = (a − b)(a² + ab + b²)
Given:
a³ − b³ = 56
and
a − b = 2
Therefore,
2(a² + ab + b²) = 56
a² + ab + b² = 28 … (1)
Also,
(a − b)² = a² − 2ab + b²
Given:
a − b = 2
Therefore,
a² − 2ab + b² = 4 … (2)
Subtracting equation (2) from equation (1):
(a² + ab + b²) − (a² − 2ab + b²) = 28 − 4
3ab = 24
ab = 8
From equation (1):
a² + b² + ab = 28
a² + b² + 8 = 28
a² + b² = 20

Question 8: If 4a² − 4a + 1 = 0 then find the value of 16a² + 1/a²

Given: 4a² − 4a + 1 = 0
or, (2a − 1)² = 0
or, 2a = 1
a = 1/2
Now,
16a² + 1/a²
= 16 × (1/2)² + 1/(1/2)²
= 16 × 1/4 + 1/(1/4)
= 4 + 4
= 8

Question 9: If (a − b + 4)² + (b + c − 9)² = 0 then find the value of a + c

Since the sum of two squares is 0, each square must be 0.
Therefore,
a − b + 4 = 0
a − b = −4 … (1)
and
b + c − 9 = 0
b + c = 9 … (2)
Adding equations (1) and (2):
(a − b) + (b + c) = −4 + 9
a + c = 5

Question 10: Simplify (p + q + r − s)² − (p − q + r + s)²

Using the identity:
A² − B² = (A + B)(A − B)
Let,
A = p + q + r − s
B = p − q + r + s
Then,
A + B
= (p + q + r − s) + (p − q + r + s)
= 2p + 2r
= 2(p + r)
Also,
A − B
= (p + q + r − s) − (p − q + r + s)
= 2q − 2s
= 2(q − s)
Therefore,
(A² − B²)
= (2(p + r))(2(q − s))
= 4(p + r)(q − s)

Question 11: If a⁴ + 1/a⁴ = 527 then find the value of a + 1/a

Using the identity:
(a² + 1/a²)² = a⁴ + 1/a⁴ + 2
Substituting the given value:
(a² + 1/a²)² = 527 + 2
(a² + 1/a²)² = 529
a² + 1/a² = 23
Now using the identity:
(a + 1/a)² = a² + 1/a² + 2
Substituting the value:
(a + 1/a)² = 23 + 2
(a + 1/a)² = 25
a + 1/a = 5

Question 12: If a² − b² + c² = 0 then find the value of (a⁶ − b⁶ + c⁶)/3(a² + c²)

Using the identity:
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
Substituting the given values:
(a + b + c)² = 34 + 2(9)
(a + b + c)² = 52
Therefore,
a + b + c = √52
Now,
(a − b)² + (b − c)² + (c − a)²
= 2(a² + b² + c² − ab − bc − ca)
Substituting the values:
= 2(34 − 9)
= 2 × 25
= 50
Therefore,
1/2[(a + b + c){(a − b)² + (b − c)² + (c − a)²}]
= 1/2 × √52 × 50
= 25√52
= 50√13

Question 13: If a² + b² + c² = 34 and ab + bc + ca = 9 then find the value of 1/2[(a + b + c){(a − b)² + (b − c)² + (c − a)²}]

For divisibility by 11, the difference between the sum of the digits in alternate positions must be 0 or a multiple of 11.
Sum of digits in odd positions: = 4 + 3 + 3 + 8 = 18
Sum of digits in even positions:
= 6 + 9 + Y
= 15 + Y
Now, because: The maximum possible difference here is between 18 and 24 (since Y is at most 9), the difference can only lie between −6 and 3.
It can never be ±11.
Therefore, the only possible value is 0.
Therefore, 18 − (15 + Y) = 0
3 − Y = 0
∴ Y = 3

Question 14: If a² + 1 = 3a then find the value of (a⁴ + 1/a²) ÷ (a² + 3a + 1)

Given: a² + 1 = 3a
Dividing both sides by a:
a + 1/a = 3 …(1)
Cubing both sides:
(a + 1/a)³ = 3³
a³ + 1/a³ + 3 × a × 1/a × (a + 1/a) = 27
a³ + 1/a³ + 3 × 1 × 3 = 27
a³ + 1/a³ = 18 …(2)
Now,
(a⁴ + 1/a²)/[a² + 3a + 1]
= a(a³ + 1/a³)/[a² + 3a + 1]
= a(a³ + 1/a³)/[(a² + 1) + 3a]
From equations (1) and (2) we get
= a(18)/[(3a) + 3a]
= 18a/6a
= 3

Question 15: If a/b + b/a = −1 then find the value of a³ − b³

Given: a/b + b/a = −1
Taking LCM:
(a² + b²)/ab = −1
Therefore,
a² + b² = −ab
Rearranging:
a² + ab + b² = 0 …(1)
Now
a³ − b³ = (a − b)(a² + ab + b²)
From equation (1):
a² + ab + b² = 0
Therefore,
a³ − b³ = (a − b) × 0
a³ − b³ = 0

Question 16: If 3a² + 2a − 6 = 0 then find the value of 27a⁶ + 54a⁵ + 36a⁴ + 8a³ − 6

Given: 3a² + 2a − 6 = 0
Therefore,
3a² + 2a = 6 …(1)
Now observe:
27a⁶ + 54a⁵ + 36a⁴ + 8a³
= (3a² + 2a)³
Using (1):
= 6³
= 216
Therefore,
27a⁶ + 54a⁵ + 36a⁴ + 8a³ − 6
= 216 − 6
= 210

Question 17: Find the value of 248³ − 300³ + 52³

Question 18: If a + 1/a = 4 then find the value of a⁶ + a⁴ − 52a³ − 14a²

Given: a + 1/a = 4 …(1)
Squaring both sides:
(a + 1/a)² = 4²
a² + 1/a² + 2 × a × 1/a = 16
a² + 1/a² = 14 …(2)
Cubing both sides of (1):
(a + 1/a)³ = 4³
a³ + 1/a³ + 3 × a × 1/a × (a + 1/a) = 64
a³ + 1/a³ + 3 × 1 × 4 = 64
a³ + 1/a³ = 52 …(3)
Now,
a⁶ + a⁴ − 52a³ − 14a²
Substituting values from (2) and (3):
= a⁶ + a⁴ − (a³ + 1/a³)a³ − (a² + 1/a²)a²
= a⁶ + a⁴ − (a⁶ + 1) − (a⁴ + 1)
= a⁶ + a⁴ − a⁶ − 1 − a⁴ − 1
= −2

Question 19: If a + b + c = 9, ab + bc + ca = 23, a³ + b³ = 35, b³ + c³ = 65, c³ + a³ = 44, then find the value of 3abc

Adding the three given cube equations:
(a³ + b³) + (b³ + c³) + (c³ + a³) = 35 + 65 + 44
2(a³ + b³ + c³) = 144
Therefore,
a³ + b³ + c³ = 72 …(1)
Using the identity:
a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
Now,
a² + b² + c² = (a + b + c)² − 2(ab + bc + ca)
Substituting the values:
a² + b² + c² = 9² − 2(23)
= 81 − 46
= 35
Therefore,
a² + b² + c² − ab − bc − ca
= 35 − 23
= 12
Using the identity:
a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
72 − 3abc = 9 × 12
72 − 3abc = 108
3abc = −36

Question 20: If x = √(7 + 4√3) then find the value of x − 1/x

Given: x = √(7 + 4√3)
Squaring:
x² = 7 + 4√3
Now,
1/x² = 1/(7 + 4√3)
Rationalising:
1/x² = (7 − 4√3)/(49 − 48)
1/x² = 7 − 4√3
Therefore,
x² + 1/x²
= (7 + 4√3) + (7 − 4√3)
= 14
Using:
(x − 1/x)² = x² + 1/x² − 2
= 14 − 2
= 12
Therefore,
x − 1/x = √12
= 2√3

Common Mistakes

Using the wrong identity for a given expression: Carefully identify the structure of the expression before applying an identity. A small change in signs or terms can require a completely different identity.
Expanding expressions unnecessarily: Many algebraic identity problems are designed to be solved by recognising patterns. Avoid lengthy expansion when a suitable identity can simplify the expression immediately.
Forgetting the sign while applying identities: In identities such as (a − b)³ or a³ − b³, incorrect handling of negative terms is a common source of errors.
Confusing (a + b)² with a² + b²: Remember that (a + b)² = a² + b² + 2ab. The middle term cannot be ignored.
Ignoring the relationship between given conditions and required expressions: Values such as a + 1/a or a² + b² are often provided to help derive higher powers. Always look for a connection before calculating.
Substituting values incorrectly after deriving an identity: When replacing expressions such as a² + 1/a² or a³ + 1/a³, ensure that the entire expression is substituted and signs are maintained correctly.
Applying cube identities incorrectly: While using a³ + b³ or a³ − b³ identities, remember the correct factorisation: a³ + b³ = (a + b)(a² − ab + b²) and a³ − b³ = (a − b)(a² + ab + b²).
Missing hidden patterns in complicated expressions: Expressions involving large powers are often constructed using identities. Look for terms such as (a + b)³, (a² + b²)², or x + 1/x before expanding.
Making assumptions about variables without sufficient information: Do not assume values of variables unless they are derived from the given conditions or identities.
Forgetting that symmetric expressions can be simplified: Expressions involving a² + b², ab + bc + ca, or a³ + b³ + c³ often require symmetric identities rather than direct calculation.

Practice Questions

Question 1: If a + b + c = 9 and ab + bc + ca = 20, find the value of a² + b² + c²

Question 2: If a² + b² = 34 and a − b = 3 find the value of ab

Question 3: If x⁴ + 1/x⁴ = 47 find the value of x + 1/x

Question 4: If a² + 1 = 4a find the value of (a⁴ + 1/a²) ÷ (a² + 4a + 1)

Question 5: If a³ + b³ + c³ = 99 a + b + c = 7 a² + b² + c² − ab − bc − ca = 5 find the value of 3abc