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 p is 321.

• A. True
• B. False

Answer 1

The correct option is B False

$a^3$- $b^3$ = $(a-b)$( $a^2$+$ab$+ $b^2$)

$a^3$+ $b^3$ =$(a+b)$( $a^2$-$ab$+ $b^2$)

Puting the values of a and b, we get the value of p as 351.

Alternatively, we can observe that since a and b are positive, so denominator has to be greater than numerator. Thus, $a^3$+ $b^3$ will be greater than $a^3$- $b^3$