If $A^{2x}=4$ what is ${A^{3x}-A^{-3x}\over A^x - A^{-x}}$

Question

If $A^{2x}=4$ and $A > 0$, what is the numerical value of $${A^{3x}-A^{-3x}\over A^x - A^{-x}}$$

Could anyone find the solution and answer? Thanks

Answer 1

Hint: Multiply numerator and denominator by $A^x$.

$$\begin{align} \frac{A^{3x}-A^{-3x}}{A^x-A^{-x}}\frac{A^x}{A^x} & = \frac{A^{4x}-A^{-2x}}{A^{2x}-1}\\ & = \frac{(A^{2x})^2-1/A^{2x}}{A^{2x}-1}\\ & = \frac{4^2-1/4}{4-1}\\ & = \frac{16-1/4}{3}\\ & = \frac{64-1}{12}\\ & = \frac{63}{12} \end{align}$$

Answer 2

Hint: $\frac{u^3-v^3}{u-v}=u^2+uv+v^2$. Let $u=A^x$ and $v=A^{-x}$.

Answer 3

Two ways:

1)$A^{2x}=4$ and $A>0$ implies $A^x=2$. Then ${A^{3x}-A^{-3x}\over A^x - A^{-x}}=\dfrac{8-\frac{1}{8}}{2-\frac{1}{2}}=\frac{21}{4}$

2)${A^{3x}-A^{-3x}\over A^x - A^{-x}}=\dfrac{\frac{A^{6x}-1}{A^{3x}}}{\frac{A^{2x}-1}{A^{x}}}=\dfrac{\frac{(A^{2x})^3-1}{A^{3x}}}{\frac{A^{2x}-1}{A^{x}}}=\dfrac{[(A^{2x})^3-1]A^{x}}{A^{3x}[A^{2x}-1]}=\dfrac{(A^{2x})^3-1}{A^{2x}[A^{2x}-1]}=\dfrac{4^3-1}{4(4-1)}=\dfrac{63}{12}=\frac{21}{4}$

Answer 4

$A^{2x} = (A^x)^2 = 4 \implies A^x = \pm 2$

${A^{3x}-A^{-3x}\over A^x - A^{-x}}= {\pm 8 - {\pm 1 \over 8}\over \pm 2 - {\pm 1 \over 2}} = {64 - 1 \over 16 - 4} = {63 \over 12}$