Question

Find the value of p, if abp = (3a +b)2 - (3a - b)2​

Answer 1

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

$abp=(9a^2+6ab+b^2)-(9a^2-6ab+b^2)$

$abp=12ab$

₁ When $ab=0$

$0p=0$

∴ Infinite solution

₂ When $ab\neq 0$

$\frac{1}{ab} \times abp=12ab \times \frac{1}{ab}$

∴ $p=12$