Simplify $\left(\dfrac{a^6 b^3}{8}\right)^{\tfrac{1}{3}}\times\dfrac{3b^{-1}}{a^2}$. Answer ... Solve $\dfrac{2}{x-2}+\dfrac{3}{x+2}=0$. Answer $x=$ ... Make $q$ the subject in the equation $pq+9=p^2+3q$ and simplify your answer. Answer ...

1. Simplify the expression inside the cube root using the power of $\tfrac13$.

$\left(\dfrac{a^6b^3}{8}\right)^{\frac13}=\dfrac{(a^6)^{\frac13}(b^3)^{\frac13}}{8^{\frac13}}=\dfrac{a^2b}{2}$

2. Write the remaining factor using positive indices where possible.

$\dfrac{3b^{-1}}{a^2}=\dfrac{3}{a^2b}$

3. Multiply the simplified factors and cancel common factors.

$\dfrac{a^2b}{2}\times\dfrac{3}{a^2b}=\dfrac{3a^2b}{2a^2b}=\dfrac{3}{2}$

1. Multiply by the common denominator to remove the fractions.

$\frac{2}{x-2}+\frac{3}{x+2}=0$
Multiply throughout by $(x-2)(x+2)$:
$2(x+2)+3(x-2)=0$

2. Expand and solve the resulting linear equation.

$2x+4+3x-6=0$
$5x-2=0$
$5x=2$
$x=\frac{2}{5}$
But substituting into the original equation gives
$\frac{2}{\frac25-2}+\frac{3}{\frac25+2}=-\frac54+\frac54=0,$
so the solution is valid.

1. Collect the terms involving $q$ on one side and the other terms on the other side.

$pq+9=p^2+3q$

$pq-3q=p^2-9$

2. Factorise the common factor $q$ from the left-hand side.

$q(p-3)=p^2-9$

3. Divide by the coefficient of $q$ and simplify the algebraic fraction.

$q=\dfrac{p^2-9}{p-3}$

$q=\dfrac{(p-3)(p+3)}{p-3}=p+3$