Exam practice

$3^{5 + (-5)} = 3^0 =$ 1.

$9^{4 - 3} = 9^1 =$ 9.

$5^{3 - 2} =$ 5.

$(4^3)^{2/3} = 4^2 =$ 16.

$27^{1/3} = (3^3)^{1/3} =$ 3.

$\left(5^{-3}\right)^{1/3} = 5^{-1} =$ $\dfrac15$.

$\left(\left(\dfrac34\right)^3\right)^{-2/3} = \left(\dfrac34\right)^{-2} = \left(\dfrac43\right)^2 =$ $\dfrac{16}{9}$.

$\left(\dfrac{216}{125}\right)^{-2/3} = \left(\dfrac65\right)^{-2} = \left(\dfrac56\right)^2 =$ $\dfrac{25}{36}$.

$(2^{-7})^{1/7} + (2^{-6})^{1/6} = 2^{-1} + 2^{-1} = \dfrac12 + \dfrac12 =$ 1.

$(2^{-4})^{1/4} \times (2^3)^{4/3} = 2^{-1} \times 2^4 = 2^3 =$ 8.

$3^3 + 3^1 + 2^3 + 1 = 27 + 3 + 8 + 1 =$ 39.

$125m^3 \div 27b^{-3} = \dfrac{125m^3b^3}{27} = \left(\dfrac{5mb}{3}\right)^3$, so the expression is $\left(\dfrac{5mb}{3}\right)^{-2} =$ $\dfrac{9}{25m^2b^2}$.

$\dfrac{64a^3b^3}{125} = \left(\dfrac{4ab}{5}\right)^3$, so the expression is $\left(\dfrac{4ab}{5}\right)^{-2} =$ $\dfrac{25}{16a^2b^2}$.

$x^{a - b + b - c + c - a} = x^0 =$ 1.

$x^{b^2 - c^2 + c^2 - a^2 + a^2 - b^2} = x^0 =$ 1.

$x^{ab + bc + ca - ab - bc - ca} = x^0 =$ 1.

$x^{(2m + 3n) + (3m + 6n) - (2m + 3n) - (4m - 4n)} = x^{3m + 6n - 4m + 4n} =$ $x^{10n - m}$.

$a^{(9n + 3 - 4n) - (2n + 10 + 3n - 7)} = a^{(5n + 3) - (5n + 3)} = a^0 =$ 1.

$7^{(n + 1 + n^2 - 1) - (n^2 - n + 2n + 2)} = 7^{(n^2 + n) - (n^2 + n + 2)} = 7^{-2} =$ $\dfrac{1}{49}$.

$\left(\dfrac{27}{8}\right)^{-1/3} = \left(\dfrac32\right)^{-1} = \dfrac23$; $\left(\dfrac{81}{16}\right)^{1/4} = \dfrac32$; $\left(\dfrac{4}{25}\right)^{-1/2} = \left(\dfrac25\right)^{-1} = \dfrac52$.
So $\dfrac23 \times \left(\dfrac32 \div \dfrac52\right) = \dfrac23 \times \dfrac35 =$ $\dfrac25$.

$\left(\dfrac{25}{16}\right)^{-1/2} = \dfrac45$; $\left(\dfrac{125}{64}\right)^{1/3} = \dfrac54$; $\left(\dfrac{8}{27}\right)^{-1/3} = \dfrac32$.
So $\dfrac45 \times \left(\dfrac54 \div \dfrac32\right) = \dfrac45 \times \dfrac56 =$ $\dfrac23$.

$\dfrac32 \times \left(\dfrac32 \div \dfrac94\right) = \dfrac32 \times \dfrac23 =$ 1 (as in Example 2a).

Numerator $3^4 \times 3^9 \times 3^{10} = 3^{23}$; denominator $3^{24} \times 3^3 \times 3^{-4} = 3^{23}$. The value is $3^0 =$ 1.

Numerator $2^4 \times 2^9 \times 2^{10} = 2^{23}$; denominator $2^{24} \times 2^3 \times 2^{-4} = 2^{23}$. The value is 1.

Multiply the first fraction’s top and bottom by $x^b$, the second’s by $x^a$:
$$\frac{x^b}{x^b + x^a} + \frac{x^a}{x^a + x^b} = \frac{x^a + x^b}{x^a + x^b}$$
1.

Multiply the first by $\dfrac{a^y}{a^y}$, the second by $\dfrac{a^x}{a^x}$:
$$\frac{a^y}{a^y - a^x} + \frac{a^x}{a^x - a^y} = \frac{a^y - a^x}{a^y - a^x}$$
1 (for x ≠ y).

Numerator $x^{2a + 2b + 2b + 2c + 2c + 2a} = x^{4(a + b + c)}$; denominator $x^{4(a + b + c)}$. 1.

Numerator $a^{3(2x + 2y + 2z)} = a^{6(x + y + z)}$, the same as the denominator. 1.

$27^{3a + 1} = 3^{9a + 3}$, $243^{-4a/5} = 3^{-4a}$, $9^{a + 5} = 3^{2a + 10}$. So
$$\begin{aligned} &3^{(9a + 3 - 4a) - (2a + 10 + 3a - 7)} \\ &= 3^{(5a + 3) - (5a + 3)} = 3^0 \end{aligned}$$
1.

$32^{2n/5} = 2^{2n}$, $4^{n + 1} = 2^{2n + 2}$. So
$$2^{(2n + 2n + 1) - (2n + 2 + 2n - 4)} = 2^{1 - (-2)} = 2^3$$
8.

$$\begin{aligned} &x^{(b + a)(b - a)} \times x^{(a + c)(a - c)} \times x^{(c + b)(c - b)} \\ &= x^{b^2 - a^2 + a^2 - c^2 + c^2 - b^2} = x^0 \end{aligned}$$
1.

$$\begin{aligned} &x^{(a - b)(a + b)} \times x^{(b - c)(b + c)} \times x^{(c - a)(c + a)} \\ &= x^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2} = x^0 \end{aligned}$$
1.

With $(a + b)(a^2 - ab + b^2) = a^3 + b^3$:
$$\begin{aligned} &x^{(m + n)(m^2 - mn + n^2)} \times x^{(n + l)(n^2 - nl + l^2)} \\ &\quad \times x^{(l + m)(l^2 - lm + m^2)} \\ &= x^{m^3 + n^3 + n^3 + l^3 + l^3 + m^3} \end{aligned}$$
$x^{2(l^3 + m^3 + n^3)}$.

The index is (a + b − c)(a − b) + (b + c − a)(b − c) + (c + a − b)(c − a). The first part is $a^2 - b^2 - ca + bc$, the second $b^2 - c^2 - ab + ca$, the third $c^2 - a^2 - bc + ab$. They add to 0, so the answer is $x^0 =$ 1.

$\sqrt[a + b]{x^{a^2 - b^2}} = x^{(a - b)(a + b)/(a + b)} = x^{a - b}$, and likewise $x^{b - c}$, $x^{c - a}$. So $x^{a - b + b - c + c - a} = x^0 =$ 1.

$$\begin{aligned} &x^{(a^2 + ab + b^2)(a - b)} \times x^{(b^2 + bc + c^2)(b - c)} \\ &\quad \times x^{(c^2 + ca + a^2)(c - a)} \\ &= x^{a^3 - b^3 + b^3 - c^3 + c^3 - a^3} = x^0 \end{aligned}$$
1.

$\dfrac{b - c}{bc} = \dfrac1c - \dfrac1b$, $\dfrac{c - a}{ca} = \dfrac1a - \dfrac1c$, $\dfrac{a - b}{ab} = \dfrac1b - \dfrac1a$. They add to 0, so the answer is $x^0 =$ 1.

$$\begin{aligned} &a^{(x + y)(x - y)} \times a^{(y - z)(y + z)} \times a^{(z - x)(z + x)} \\ &= a^{x^2 - y^2 + y^2 - z^2 + z^2 - x^2} = a^0 \end{aligned}$$
1.

LHS $= x^{(b - c)/(bc)} \times x^{(c - a)/(ca)} \times x^{(a - b)/(ab)} = x^{(1/c - 1/b) + (1/a - 1/c) + (1/b - 1/a)} = x^0 = 1 =$ RHS.

LHS $= x^{(a - b)/2} \times x^{(b - c)/2} \times x^{(c - a)/2} = x^{(a - b + b - c + c - a)/2} = x^0 = 1 =$ RHS.

This is Example 4a. The factors are $x^{1/c^2 - 1/b^2}$, $x^{1/a^2 - 1/c^2}$ and $x^{1/b^2 - 1/a^2}$; their indices add to 0, so LHS $= x^0 = 1 =$ RHS.

Multiply the top and bottom of the first fraction by $x^b$, the second by $x^c$, the third by $x^a$:
$$\begin{aligned} &\frac{x^b}{x^b + x^a + x^c} + \frac{x^c}{x^c + x^b + x^a} \\ &\quad + \frac{x^a}{x^a + x^c + x^b} \end{aligned}$$
$= \dfrac{x^a + x^b + x^c}{x^a + x^b + x^c} = 1 =$ RHS.

$\dfrac{p + \frac1q}{q + \frac1p} = \dfrac{pq + 1}{q} \times \dfrac{p}{pq + 1} = \dfrac pq$ and $\dfrac{p - \frac1q}{q - \frac1p} = \dfrac{pq - 1}{q} \times \dfrac{p}{pq - 1} = \dfrac pq$. So LHS $= \left(\dfrac pq\right)^m \left(\dfrac pq\right)^m = \left(\dfrac pq\right)^{2m} =$ RHS.

As in Example 5, with x, y for p, q and a, b for the indices. Numerator: $\left(x + \frac1y\right)^a\left(x - \frac1y\right)^a\left(x - \frac1y\right)^{b - a} = \left(x + \frac1y\right)^a\left(x - \frac1y\right)^b$. Denominator: $\left(y + \frac1x\right)^b\left(y - \frac1x\right)^b\left(y + \frac1x\right)^{a - b} = \left(y + \frac1x\right)^a\left(y - \frac1x\right)^b$.
Since $\dfrac{x + \frac1y}{y + \frac1x} = \dfrac{xy + 1}{y} \times \dfrac{x}{xy + 1} = \dfrac xy$ and $\dfrac{x - \frac1y}{y - \frac1x} = \dfrac xy$, LHS $= \left(\dfrac xy\right)^a\left(\dfrac xy\right)^b = \left(\dfrac xy\right)^{a + b} =$ RHS.

This is Example 5 with a, b for p, q. The numerator is $\left(a + \frac1b\right)^x\left(a - \frac1b\right)^y$, the denominator $\left(b + \frac1a\right)^x\left(b - \frac1a\right)^y$, and each ratio of brackets is $\dfrac ab$. So LHS $= \left(\dfrac ab\right)^{x + y} =$ RHS.

Each bracket is $x^{p + q}$, $x^{q + r}$, $x^{r + p}$, and $(p + q)(p^2 - pq + q^2) = p^3 + q^3$. So
$$\begin{aligned} \text{LHS} &= x^{p^3 + q^3 + q^3 + r^3 + r^3 + p^3} \\ &= x^{2(p^3 + q^3 + r^3)} = x^{2 \times 1} = x^2 \end{aligned}$$
= RHS.

$$\begin{aligned} \text{LHS} &= a^{(x - y)^2 + (y - z)^2 + (z - x)^2} \\ &= a^{2(x^2 + y^2 + z^2) - 2(xy + yz + zx)} \end{aligned}$$
By the condition the two brackets in the index are equal, so the index is 0 and LHS $= a^0 = 1 =$ RHS.

From the condition, $m^{-f} = m^{g + h}$ and $m^{g + h + f} = 1$. Multiply the first fraction’s top and bottom by $m^h$ and the third’s by $m^{g + h}$:
$$\begin{aligned} &\frac{m^h}{m^h + m^{g + h} + 1} + \frac{1}{1 + m^h + m^{g + h}} \\ &\quad + \frac{m^{g + h}}{m^{g + h} + 1 + m^h} \end{aligned}$$
All three denominators are $1 + m^h + m^{g + h}$, and so is the sum of the numerators. LHS = 1 = RHS.

As in Example 6: $x^{-t} = x^{r + s}$ and $x^{r + s + t} = 1$. Multiply the first fraction by $\dfrac{x^s}{x^s}$ and the third by $\dfrac{x^{r + s}}{x^{r + s}}$:
$$\begin{aligned} &\frac{x^s}{x^s + x^{r + s} + 1} + \frac{1}{1 + x^s + x^{r + s}} \\ &\quad + \frac{x^{r + s}}{x^{r + s} + 1 + x^s} \end{aligned}$$
The numerators add up to the common denominator, so LHS = 1 = RHS.

From xyz = 1, $z^{-1} = xy$. Aim for the denominator 1 + y + xy in every term.

First term, top and bottom × y: $\dfrac{y}{y + xy + 1}$.
Second term, with $z^{-1} = xy$: $\dfrac{1}{1 + y + xy}$.
Third term, top and bottom × xy: $\dfrac{xy}{xy + xyz + y} = \dfrac{xy}{xy + 1 + y}$.
LHS $= \dfrac{y + 1 + xy}{1 + y + xy} = 1 =$ RHS.

$(6a)^3 + \left(\dfrac12\right)^3 =$ $\left(6a + \dfrac12\right)\left(36a^2 - 3a + \dfrac14\right)$, by $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.

$(2x)^3 - (5y)^3 =$ $(2x - 5y)(4x^2 + 10xy + 25y^2)$.

A cube is already a product: (x + 2)(x + 2)(x + 2). Multiplied out it is $x^3 + 6x^2 + 12x + 8$, so if a question gives that, these are its factors.

(x − 3)(x − 3)(x − 3); multiplied out, $x^3 - 9x^2 + 27x - 27$.

$$\begin{aligned} &(4m^2 - 7n^2)^2 + 56m^2n^2 - 65m^2n^2 \\ &= (4m^2 - 7n^2)^2 - (3mn)^2 \\ &= (4m^2 + 3mn - 7n^2) \\ &\quad \times (4m^2 - 3mn - 7n^2) \end{aligned}$$
Each factor splits further: $4m^2 + 3mn - 7n^2 = (4m + 7n)(m - n)$ and $4m^2 - 3mn - 7n^2 = (4m - 7n)(m + n)$. Fully factorized: (m + n)(m − n)(4m + 7n)(4m − 7n). (Quicker: it is $(16m^2 - 49n^2)(m^2 - n^2)$.)

$(3x^2)^2 - 2 \times 3x^2 \times \dfrac{1}{3y^2} + \left(\dfrac{1}{3y^2}\right)^2 =$ $\left(3x^2 - \dfrac{1}{3y^2}\right)\left(3x^2 - \dfrac{1}{3y^2}\right)$.
Each bracket is also a difference of two squares, $3x^2 - \dfrac{1}{3y^2} = 3\left(x^2 - \dfrac{1}{9y^2}\right) = 3\left(x + \dfrac{1}{3y}\right)\left(x - \dfrac{1}{3y}\right)$, so taken all the way it is $9\left(x + \dfrac{1}{3y}\right)^2\left(x - \dfrac{1}{3y}\right)^2$.