Simplify $(2x^3y^2)^{-4}\;\div\;(10x^{-2}y^3)^2\;\div\;\sqrt[3]{27x^{-3}y^6}.$ Answer: ... Solve $3^{x+2}\times 3(3^5) = 1.$

1. Simplify each bracketed or rooted factor separately.

$(2x^3y^2)^{-4}=2^{-4}x^{-12}y^{-8}=\frac{x^{-12}y^{-8}}{16}$
$(10x^{-2}y^3)^2=100x^{-4}y^6$
$\sqrt[3]{27x^{-3}y^6}=3x^{-1}y^2$

2. Write the full expression using the simplified factors and combine numerical constants.

$\frac{x^{-12}y^{-8}}{16}\div 100x^{-4}y^6\div 3x^{-1}y^2$
$=\frac{1}{16\times100\times3}\,x^{-12-(-4)-(-1)}y^{-8-6-2}$

3. Simplify the powers of $x$ and $y$ and the constant.

$\frac{1}{16\times100\times3}\,x^{-7}y^{-16}$
$=\frac{x^{-7}y^{-16}}{4800}$
$=\frac{1}{4800x^7y^{16}}$

1. Write all factors as powers of 3 and combine the exponents.

$3^{x+2}\times 3(3^5)=1$

Since $3=3^1$,
$3^{x+2}\times 3^1\times 3^5=1$
$3^{x+2+1+5}=1$
$3^{x+8}=1$

2. Use $1=3^0$ and solve for $x$.

$3^{x+8}=3^0$
Equating indices,
$x+8=0$
$x=-8$