Question

Prove the following identities :

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

$\left(ii\right){(a-b)}^3=a^3-b^3-3a^{2}b+3b^{2}a$ [4 MARKS]

Answer 1

Application: 2 Marks each

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

$=(a^2+b^2+2ab)(a+b)$

$=a^3+a^{2}b+a{b}^2+b^3+2a^{2}b+2a{b}^2$

$=a^3+b^3+3a^{2}b+3a{b}^2$


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

$=(a^2+b^2-2ab)(a-b)$

$=a^3-a^{2}b+a{b}^2-b^3-2a^{2}b+2a{b}^2$

$=a^3-b^3-3a^{2}b+3a{b}^2$