Without actually calculating the cubes, find the value of each of the following
$(31)^{3}-(16)^{3}-(15)^{3}$
$14250$
$22500$
$21436$
$22320$
The zero of polynomial $p(x)=b x+m$ is $\ldots \ldots \ldots$
Divide $p(x)=x^{3}+7 x^{2}+14 x+1$ by $x+3$ and find the quotient and the remainder.
Multiply $x^{2}+4 y^{2}+z^{2}+2 x y+x z-2 y z$ by $(-z+x-2 y)$
On dividing $16 x^{2}-24 x+9$ by $4 x-3,$ find the remainder.
Factorise :
$a^{3}-8 b^{3}-64 c^{3}-24 a b c$
Confusing about what to choose? Our team will schedule a demo shortly.
