Question

Show that $\frac{4^3-1}{3}=4^2+4+1$.

Answer 1

If $=\frac{4^3-1}{(4-1)}$

Ans $=\frac{4^3-13}{4-1}=\frac{(4-1)(4^2+(4\left)\right(1)+1^2)}{3}$ $\{Byformul{a}^{\prime }s{a}^3+b^3=(a-b\left)\right(a^2+b^2+ab\left)\right\}$

$=4^2+4+1$

$\frac{a^n-1}{a-1}=[a^{n-1}+a^{n-2}+---a+1]$ {By mathematics induction

$\frac{4^3-1}{4-1}=[4^{(3-1)}+4^{(3-2)}+4^{(3-3)}...]$

$=4^2+4+1$