Question

If the value of $\left(\frac{a^3-b^3}{a^3+b^3}\right)$ when a = 7 and b = 2 is $\frac{335}{p}$, then find the value of p.

Answer 1

Given that a = 7 and b = 2

$\therefore$ $a^3$- $b^3$ = $7^3-2^3$ = 343 - 8 = 335 -------(1)

and $a^3$+ $b^3$ = $7^3+2^3$ = 343 + 8 = 351-----(2)

$\frac{a^3-b^3}{a^3+b^3}$ = $\frac{335}{351}$ = $\frac{335}{p}$

Comparing , we get p = 351