Express $\frac{12xy^2}{15} \div \frac{2xy^3}{(3x)^2}$ as a single fraction in its simplest form. Given that $\dfrac{v}{3} = \sqrt[3]{\dfrac{2s+v}{s-3}}$, express $s$ in terms of $v$. Solve the equation $\dfrac{3}{2x-1}=\dfrac{5}{x-2}$. Simplify $(x^2+y^2)(x^2-y^2)$. Show that $2(9^{n+1})+3^{2n+3}-9^n$ is a multiple of 11 for all positive integer values of $n$.

1. Rewrite the divisor by simplifying $(3x)^2$ and change division to multiplication by the reciprocal.

$\dfrac{12xy^2}{15} \div \dfrac{2xy^3}{(3x)^2} = \dfrac{12xy^2}{15} \div \dfrac{2xy^3}{9x^2} = \dfrac{12xy^2}{15} \times \dfrac{9x^2}{2xy^3}$

2. Cancel common numerical and algebraic factors to obtain the simplest single fraction.

$\dfrac{12xy^2}{15} \times \dfrac{9x^2}{2xy^3} = \dfrac{12 \times 9 \times x \times x^2 \times y^2}{15 \times 2 \times x \times y^3} = \dfrac{18x^2}{5y}$

1. Cube both sides to eliminate the cube root.

$\frac{v}{3}=\sqrt[3]{\frac{2s+v}{s-3}}$
Cubing both sides,
$\left(\frac{v}{3}\right)^3=\frac{2s+v}{s-3}$
$\frac{v^3}{27}=\frac{2s+v}{s-3}$

2. Clear the denominator and collect the terms involving $s$.

$v^3(s-3)=27(2s+v)$
$v^3s-3v^3=54s+27v$
$v^3s-54s=3v^3+27v$

3. Factorise and divide to express $s$ in terms of $v$.

$s(v^3-54)=3v^3+27v$
$s=\frac{3v^3+27v}{v^3-54}$
$s=\frac{3v(v^2+9)}{v^3-54}$

Equivalently, leaving the numerator as $3v^3+27v$ is acceptable.

1. State the restrictions and cross-multiply to remove the fractions.

The denominators must not be zero, so $x\ne \dfrac12$ and $x\ne 2$.

$\frac{3}{2x-1}=\frac{5}{x-2}$
$3(x-2)=5(2x-1)$

2. Expand and solve the resulting linear equation.

$3x-6=10x-5$
$-1=7x$
$x=-\frac17$

1. Apply the difference of two squares identity.

$(x^2+y^2)(x^2-y^2)=(x^2)^2-(y^2)^2=x^4-y^4$

1. Rewrite the powers so that each term is expressed in terms of $9^n$.

Since $9^{n+1}=9\cdot 9^n$ and
$3^{2n+3}=3^{2n}\cdot 3^3=(3^2)^n\cdot 27=27\cdot 9^n,$
we have
$2(9^{n+1})+3^{2n+3}-9^n=2(9\cdot 9^n)+27\cdot 9^n-9^n.$

2. Collect like terms involving $9^n$.

$2(9\cdot 9^n)+27\cdot 9^n-9^n=18\cdot 9^n+27\cdot 9^n-9^n$
$=(18+27-1)9^n=44\cdot 9^n.$

3. Factor out 11 to prove the expression is a multiple of 11.

$44\cdot 9^n=11(4\cdot 9^n).$
Since $n$ is a positive integer, $4\cdot 9^n$ is an integer. Hence $11(4\cdot 9^n)$ is a multiple of $11$.