Expand each of the following, using suitable identities : $\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$
$\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$
Using $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x,$ we have
$\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}=\left(\frac{1}{4} a\right)^{2}+\left(-\frac{1}{2} b\right)^{2}+(1)^{2}+2\left(\frac{1}{4} a\right)\left(-\frac{1}{2} b\right)+2\left(-\frac{1}{2} b\right)(1)+2(1)\left(\frac{1}{4} a\right)$
$\quad=\frac{1}{16} a^{2}+\frac{1}{4} b^{2}+1+\left[-\frac{1}{4} a b\right]+[-b]+\left[\frac{1}{2} a\right]$
$\quad=\frac{1}{16} a^{2}+\frac{1}{4} b^{2}+1-\frac{1}{4} a b-b+\frac{1}{2} a$
Expand each of the following, using suitable identities : $(2 x-y+z)^{2}$
Without actually calculating the cubes, find the value of each of the following : $(-12)^{3}+(7)^{3}+(5)^{3}$
Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases : $p(x)=x^{3}+3 x^{2}+3 x+1$, $g(x)=x+2$.
Factorise : $x^{3}-2 x^{2}-x+2$
Find the remainder when $x^{3}+3 x^{2}+3 x+1$ is divided by $x$.