On dividing $p(x)=2 x^{3}-3 x^{2}+a x-3 a+9$ by $(x+1),$ if the remainder is $16,$ then find the value of $a$. Then, find the remainder on dividing $p(x)$ by $x+2$
Without actually calculating the cubes, find the value of :
$(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$
Factorise
$6 x^{3}-23 x^{2}+29 x-12$
Expand
$\left(\frac{2 x}{3}+\frac{4 y}{5}\right)\left(\frac{2 x}{3}-\frac{4 y}{5}\right)$
Factorise
$25 x^{2}+25 x+6$