(a-b) ^3 formula – Derivation – solved examples – FAQ

(a-b)^3 formula

The formula for the expansion of (a-b)^3 is \left(a^{3}+3 a b^{2}-3 a^{2} b-b^{3}\right).

It is the formula for the cube of the difference between two numbers.

Derivation

We can write (a-b)^{3} as the product of (a-b) multiplied with itself three times.

(a-b)^{3}=(a-b)(a-b)(a-b)

\rightarrow(a-b)^{3}=\left(a^{2}+b^{2}-2 a b\right)(a-b)

\rightarrow(a-b)^{3}=\left(a^{3}+a b^{2}-2 a^{2} b-a^{2} b-b^{3}+2 a b^{2}\right)

\rightarrow(a-b)^{3}=\left(a^{3}+3 a b^{2}-3 a^{2} b-b^{3}\right)

Hence the expansion of (a-b)^{3} \text { is }\left(a^{3}+3 a b^{2}-3 a^{2} b-b^{3}\right)

Check

Let us take a=9and  b=5

\mathrm{LHS}=(a-b)^{3}

=(9-5)^{3}

=4^{3}

=64

\text { RHS }=a^{3}+3 a b^{2}-3 a^{2} b-b^{3}

=9^{3}+3 \times 9 \times 5^{2}-3 \times 9^{2} \times 5-5^{3}

=729+675-1215-125

LHS = RHS

\text { Hence }(a-b)^{3}=\left(a^{3}+3 a b^{2}-3 a^{2} b-b^{3}\right)

Physical representation

Suppose there is a cube of side “a” cm. If we shorten each side by “b” cm, then the volume of the resultant cube is equal to(a-b)^{3}

Solved Examples

\text { 1. Expand }(2 a-b)^{3}

(2 a-b)^{3} =\left((2 a)^{3}+3(2 a) b^{2}-3(2 a)^{2} b-b^{3}\right)

=8 a^{3}+6 a b^{2}-12 a^{2} b-b^{3}

2. Find the value of \left(m^{3}-n^{3}\right)if m-n = 5 and mn=36

(m-n)^{3} =\left(m^{3}+3 m n^{2}-3 m^{2} n-n^{3}\right)

\rightarrow m^{3}-n^{3}=(5)^{3}+3 \times 36 \times 5

Hence \left(m^{3}-n^{3}\right)is equal to 665

3. Expand (a-3 b)^{3}+(4 a-b)^{3}

(a-3 b)^{3}=\left((a)^{3}+3(a)(3 b)^{2}-3(a)^{2}(3 b)-(3 b)^{3}\right)

=a^{3}+27 a b^{2}-9 a^{2} b-9 b^{3}

(4 a-b)^{3}=\left((4 a)^{3}+3(4 a) b^{2}-3(4 a)^{2} b-b^{3}\right)

=64 a^{3}+12 a b^{2}-48 a^{2} b-b^{3}

(a-3 b)^{3}+(4 a-b)^{3}  =\left(a^{3}+27 a b^{2}-9 a^{2} b-9 b^{3}\right)+\left(64 a^{3}+12 a b^{2}-48 a^{2} b-b^{3}\right)

=65 a^{3}+29 a b^{2}-57 a^{2} b-10 b^{3}

\text { Hence, }(a-3 b)^{3}+(4 a-b)^{3} \text { is equal to } 65 a^{3}+29 a b^{2}-57 a^{2} b-10 b^{3}