Exam practice

$(x + 1)^3 = x^3 + 3 \cdot x^2 \cdot 1 + 3 \cdot x \cdot 1^2 + 1^3 =$ $x^3 + 3x^2 + 3x + 1$.

$(x - 3)^3 = x^3 - 3 \cdot x^2 \cdot 3 + 3 \cdot x \cdot 3^2 - 3^3 =$ $x^3 - 9x^2 + 27x - 27$.

$(x + 4)^3 = x^3 + 3 \cdot x^2 \cdot 4 + 3 \cdot x \cdot 4^2 + 4^3 =$ $x^3 + 12x^2 + 48x + 64$.

$(2x - 1)^3 = (2x)^3 - 3(2x)^2(1) + 3(2x)(1)^2 - 1^3 =$ $8x^3 - 12x^2 + 6x - 1$.

$(3y + 2b)^3 = (3y)^3 + 3(3y)^2(2b) + 3(3y)(2b)^2 + (2b)^3 =$ $27y^3 + 54y^2b + 36yb^2 + 8b^3$.

$$\begin{aligned} &\left(\frac{x}{2} - 1\right)^3 \\ &= \left(\frac{x}{2}\right)^3 - 3\left(\frac{x}{2}\right)^2(1) + 3\left(\frac{x}{2}\right)(1)^2 - 1^3 \\ &= \frac{x^3}{8} - \frac{3x^2}{4} + \frac{3x}{2} - 1 \end{aligned}$$
$\dfrac{1}{8}x^3 - \dfrac{3}{4}x^2 + \dfrac{3}{2}x - 1$

$(2x)^3 + 3(2x)^2(3y) + 3(2x)(3y)^2 + (3y)^3 =$ $8x^3 + 36x^2y + 54xy^2 + 27y^3$.

$(5a)^3 - 3(5a)^2(8b) + 3(5a)(8b)^2 - (8b)^3 =$ $125a^3 - 600a^2b + 960ab^2 - 512b^3$.

The first and last terms are the cubes: find what they are cubes of, then check the middle terms.

$27a^3 = (3a)^3$ and $64b^3 = (4b)^3$. Check: $3(3a)^2(4b) = 108a^2b$ ✓ and $3(3a)(4b)^2 = 144ab^2$ ✓.
$27a^3 + 108a^2b + 144ab^2 + 64b^3 = (3a + 4b)^3$

$8x^3 = (2x)^3$, $27y^3 = (3y)^3$; $3(2x)^2(3y) = 36x^2y$ ✓, $3(2x)(3y)^2 = 54xy^2$ ✓.
$(2x + 3y)^3$

$64m^3 = (4m)^3$, $n^3 = n^3$; $3(4m)^2n = 48m^2n$ ✓, $3(4m)n^2 = 12mn^2$ ✓, and the signs go + −+ −.
$(4m - n)^3$

$125p^3 = (5p)^3$, $8q^3 = (2q)^3$; $3(5p)^2(2q) = 150p^2q$ ✓, $3(5p)(2q)^2 = 60pq^2$ ✓.
$(5p - 2q)^3$

$$\begin{aligned} 8x^4 + 27x &= x(8x^3 + 27) \\ &= x\{(2x)^3 + 3^3\} \\ &= x(2x + 3) \\ &\quad \times \{(2x)^2 - 2x \cdot 3 + 3^2\} \end{aligned}$$
$x(2x + 3)(4x^2 - 6x + 9)$

$x^3 + (2y)^3 = (x + 2y)\{x^2 - x \cdot 2y + (2y)^2\} =$ $(x + 2y)(x^2 - 2xy + 4y^2)$.

$1^3 + (5x)^3 = (1 + 5x)\{1 - 5x + (5x)^2\} =$ $(1 + 5x)(1 - 5x + 25x^2)$.

With a = 2x, $b = \dfrac{1}{x}$, and ab = 2:
$$\begin{aligned} &(2x)^3 + \left(\frac{1}{x}\right)^3 \\ &= \left(2x + \frac{1}{x}\right)\left(4x^2 - 2 + \frac{1}{x^2}\right) \end{aligned}$$
$\left(2x + \dfrac{1}{x}\right)\left(4x^2 - 2 + \dfrac{1}{x^2}\right)$

Here a + b plays the part of the first cube’s base, and 1 the second’s:
$$\begin{aligned} &(a + b)^3 + 1^3 \\ &= (a + b + 1)\{(a + b)^2 - (a + b) + 1\} \end{aligned}$$
Expand $(a + b)^2$ in the second bracket:
$(a + b + 1)(a^2 + 2ab + b^2 - a - b + 1)$

$x^6 = (x^2)^3$.

$(x^2)^3 + (y^2)^3 = (x^2 + y^2)\{(x^2)^2 - x^2y^2 + (y^2)^2\} =$ $(x^2 + y^2)(x^4 - x^2y^2 + y^4)$.

$$\begin{aligned} &250m^4 - 2m \\ &= 2m(125m^3 - 1) \\ &= 2m\{(5m)^3 - 1^3\} \\ &= 2m(5m - 1)(25m^2 + 5m + 1) \end{aligned}$$
$2m(5m - 1)(25m^2 + 5m + 1)$

$$\begin{aligned} x^3y - 64y^4 &= y(x^3 - 64y^3) \\ &= y\{x^3 - (4y)^3\} \\ &= y(x - 4y)(x^2 + 4xy + 16y^2) \end{aligned}$$
$y(x - 4y)(x^2 + 4xy + 16y^2)$

$64p^6q^3 = (4p^2q)^3$ and $125 = 5^3$:
$$\begin{aligned} &(4p^2q)^3 - 5^3 \\ &= (4p^2q - 5)\{(4p^2q)^2 + 4p^2q \cdot 5 + 5^2\} \end{aligned}$$
$(4p^2q - 5)(16p^4q^2 + 20p^2q + 25)$

$8(a + b)^3 = \{2(a + b)\}^3$.

It is $A^3 - B^3$ with A = a − b and B = 2a + 2b, so it is $(A - B)(A^2 + AB + B^2)$. Work out the brackets one at a time:
$$\begin{aligned} A - B &= a - b - 2a - 2b \\ &= -(a + 3b) \end{aligned}$$
$$\begin{aligned} &A^2 + AB + B^2 \\ &= (a - b)^2 + (a - b)(2a + 2b) \\ &\quad + (2a + 2b)^2 \\ &= (a^2 - 2ab + b^2) + (2a^2 - 2b^2) \\ &\quad + (4a^2 + 8ab + 4b^2) \\ &= 7a^2 + 6ab + 3b^2 \end{aligned}$$
So $(a - b)^3 - 8(a + b)^3 = -(a + 3b)(7a^2 + 6ab + 3b^2)$.
$-(a + 3b)(7a^2 + 6ab + 3b^2)$

As in Example 4b, with $a = \dfrac{x}{y}$, $b = \dfrac{y}{x}$, ab = 1:
$$\begin{aligned} \frac{x^3}{y^3} - \frac{y^3}{x^3} &= \left(\frac{x}{y} - \frac{y}{x}\right)\left(\frac{x^2}{y^2} + 1 + \frac{y^2}{x^2}\right) \\ &= \left(\frac{x}{y} - \frac{y}{x}\right)\left\{\left(\frac{x}{y} + \frac{y}{x}\right)^2 - 1\right\} \end{aligned}$$
$\left(\dfrac{x}{y} - \dfrac{y}{x}\right)\left(\dfrac{x}{y} + \dfrac{y}{x} + 1\right)\left(\dfrac{x}{y} + \dfrac{y}{x} - 1\right)$
Stopping at the first line, $\left(\dfrac{x}{y} - \dfrac{y}{x}\right)\left(\dfrac{x^2}{y^2} + 1 + \dfrac{y^2}{x^2}\right)$, is also a correct factorization, but not a complete one.

With a = p, $b = \dfrac{1}{p}$, ab = 1:
$$\begin{aligned} p^3 - \frac{1}{p^3} &= \left(p - \frac{1}{p}\right)\left(p^2 + 1 + \frac{1}{p^2}\right) \\ &= \left(p - \frac{1}{p}\right)\left\{\left(p + \frac{1}{p}\right)^2 - 1\right\} \end{aligned}$$
$\left(p - \dfrac{1}{p}\right)\left(p + \dfrac{1}{p} + 1\right)\left(p + \dfrac{1}{p} - 1\right)$
($\left(p - \dfrac{1}{p}\right)\left(p^2 + 1 + \dfrac{1}{p^2}\right)$ is correct too, one step short.)

The log’s volume is $x^3$ ft³ and the piece cut out is $7^3 = 343$ ft³.
Wood left $= x^3 - 343 = x^3 - 7^3 =$ $(x - 7)(x^2 + 7x + 49)$ ft³, that is, $(x^3 - 343)$ ft³.
What it means: if the log is 10 ft on each edge, 3 × (100 + 70 + 49) = 657 ft³ is left, out of 1,000.

$$\begin{aligned} &x^4 + x^2y^2 + y^4 \\ &= (x^2 + y^2)^2 - 2x^2y^2 + x^2y^2 \\ &= (x^2 + y^2)^2 - (xy)^2 \end{aligned}$$
$(x^2 + xy + y^2)(x^2 - xy + y^2)$

$$\begin{aligned} 16x^4 + 7x^2 + 1 &= (4x^2 + 1)^2 - 8x^2 + 7x^2 \\ &= (4x^2 + 1)^2 - x^2 \end{aligned}$$
$(4x^2 + x + 1)(4x^2 - x + 1)$

$$\begin{aligned} &16x^4 + 36x^2y^2 + 81y^4 \\ &= (4x^2 + 9y^2)^2 - 72x^2y^2 + 36x^2y^2 \\ &= (4x^2 + 9y^2)^2 - (6xy)^2 \end{aligned}$$
$(4x^2 + 6xy + 9y^2)(4x^2 - 6xy + 9y^2)$

$$\begin{aligned} &4m^4 + 35m^2n^2 + 121n^4 \\ &= (2m^2 + 11n^2)^2 - 44m^2n^2 + 35m^2n^2 \\ &= (2m^2 + 11n^2)^2 - (3mn)^2 \end{aligned}$$
$(2m^2 + 3mn + 11n^2)(2m^2 - 3mn + 11n^2)$

Take out the common factor 3 first.

$$\begin{aligned} &48a^4 + 108a^2b^2 + 243b^4 \\ &= 3(16a^4 + 36a^2b^2 + 81b^4) \\ &= 3\{(4a^2 + 9b^2)^2 - 72a^2b^2 + 36a^2b^2\} \\ &= 3\{(4a^2 + 9b^2)^2 - (6ab)^2\} \end{aligned}$$
$3(4a^2 + 6ab + 9b^2)(4a^2 - 6ab + 9b^2)$

$$\begin{aligned} &32p^4 + 72p^2q^2 + 162q^4 \\ &= 2(16p^4 + 36p^2q^2 + 81q^4) \\ &= 2\{(4p^2 + 9q^2)^2 - (6pq)^2\} \end{aligned}$$
$2(4p^2 + 6pq + 9q^2)(4p^2 - 6pq + 9q^2)$

$$\begin{aligned} x^4 + 4 &= (x^2 + 2)^2 - 4x^2 \\ &= (x^2 + 2)^2 - (2x)^2 \end{aligned}$$
$(x^2 + 2x + 2)(x^2 - 2x + 2)$

$$\begin{aligned} 4x^4 + 81y^4 &= (2x^2 + 9y^2)^2 - 36x^2y^2 \\ &= (2x^2 + 9y^2)^2 - (6xy)^2 \end{aligned}$$
$(2x^2 + 6xy + 9y^2)(2x^2 - 6xy + 9y^2)$

$$\begin{aligned} 64e^4 + f^4 &= (8e^2 + f^2)^2 - 16e^2f^2 \\ &= (8e^2 + f^2)^2 - (4ef)^2 \end{aligned}$$
$(8e^2 + 4ef + f^2)(8e^2 - 4ef + f^2)$

$$\begin{aligned} m^4 + 4n^4 &= (m^2 + 2n^2)^2 - 4m^2n^2 \\ &= (m^2 + 2n^2)^2 - (2mn)^2 \end{aligned}$$
$(m^2 + 2mn + 2n^2)(m^2 - 2mn + 2n^2)$, Sophie Germain’s rule itself.

$$\begin{aligned} 81x^4 + 64y^4 &= (9x^2 + 8y^2)^2 - 144x^2y^2 \\ &= (9x^2 + 8y^2)^2 - (12xy)^2 \end{aligned}$$
$(9x^2 + 12xy + 8y^2)(9x^2 - 12xy + 8y^2)$

$$\begin{aligned} y^4 + 324x^4 &= (y^2 + 18x^2)^2 - 36x^2y^2 \\ &= (y^2 + 18x^2)^2 - (6xy)^2 \end{aligned}$$
$(y^2 + 6xy + 18x^2)(y^2 - 6xy + 18x^2)$

$$\begin{aligned} &x^4 - 5x^2y^2 + 4y^4 \\ &= (x^2 + 2y^2)^2 - 4x^2y^2 - 5x^2y^2 \\ &= (x^2 + 2y^2)^2 - (3xy)^2 \end{aligned}$$
$(x^2 + 3xy + 2y^2)(x^2 - 3xy + 2y^2)$
Each bracket splits again, by splitting the middle term: $x^2 + 3xy + 2y^2 = (x + y)(x + 2y)$ and $x^2 - 3xy + 2y^2 = (x - y)(x - 2y)$. So in full, (x + y)(x − y)(x + 2y)(x − 2y).

$(x^2 + 3y^2)^2$ leaves $-28x^2y^2$, and 28 is not a square. Try $(x^2 - 3y^2)^2$.

$$\begin{aligned} &x^4 - 22x^2y^2 + 9y^4 \\ &= (x^2 - 3y^2)^2 + 6x^2y^2 - 22x^2y^2 \\ &= (x^2 - 3y^2)^2 - (4xy)^2 \end{aligned}$$
$(x^2 + 4xy - 3y^2)(x^2 - 4xy - 3y^2)$

$$\begin{aligned} b^4 - 3b^2 + 1 &= (b^2 - 1)^2 + 2b^2 - 3b^2 \\ &= (b^2 - 1)^2 - b^2 \end{aligned}$$
$(b^2 + b - 1)(b^2 - b - 1)$

Here both squares work.
With $(5x^2 - 3y^2)^2$: $(5x^2 - 3y^2)^2 + 30x^2y^2 - 34x^2y^2 = (5x^2 - 3y^2)^2 - (2xy)^2$, so $(5x^2 + 2xy - 3y^2)(5x^2 - 2xy - 3y^2)$.
With $(5x^2 + 3y^2)^2$: $(5x^2 + 3y^2)^2 - 30x^2y^2 - 34x^2y^2 = (5x^2 + 3y^2)^2 - (8xy)^2$, so $(5x^2 + 8xy + 3y^2)(5x^2 - 8xy + 3y^2)$.
Two different-looking answers can both be right because neither is finished: each bracket splits again, and both end at the same four factors, (5x + 3y)(5x − 3y)(x + y)(x − y).

As in Example 3: $(7a^2 + 3b^2)^2 - 196a^2b^2 = (7a^2 + 3b^2)^2 - (14ab)^2$.
$(7a^2 + 14ab + 3b^2)(7a^2 - 14ab + 3b^2)$

$$\begin{aligned} &25a^5b - 9a^3b^3 + 16ab^5 \\ &= ab(25a^4 - 9a^2b^2 + 16b^4) \\ &= ab\{(5a^2 + 4b^2)^2 - 40a^2b^2 - 9a^2b^2\} \\ &= ab\{(5a^2 + 4b^2)^2 - (7ab)^2\} \end{aligned}$$
$ab(5a^2 + 7ab + 4b^2)(5a^2 - 7ab + 4b^2)$

The end terms are $\left(\dfrac{m^2}{n^2}\right)^2$ and $\left(\dfrac{n^2}{m^2}\right)^2$, whose product is 1:
$$\begin{aligned} \frac{m^4}{n^4} + \frac{n^4}{m^4} + 1 &= \left(\frac{m^2}{n^2} + \frac{n^2}{m^2}\right)^2 - 2 + 1 \\ &= \left(\frac{m^2}{n^2} + \frac{n^2}{m^2}\right)^2 - 1^2 \end{aligned}$$
$\left(\dfrac{m^2}{n^2} + 1 + \dfrac{n^2}{m^2}\right)\left(\dfrac{m^2}{n^2} - 1 + \dfrac{n^2}{m^2}\right)$

$$\begin{aligned} y^4 + \frac{1}{y^4} + 1 &= \left(y^2 + \frac{1}{y^2}\right)^2 - 2 + 1 \\ &= \left(y^2 + \frac{1}{y^2}\right)^2 - 1^2 \end{aligned}$$
$\left(y^2 + 1 + \dfrac{1}{y^2}\right)\left(y^2 - 1 + \dfrac{1}{y^2}\right)$

$$\begin{aligned} &\frac{a^4}{b^4} - \frac{5a^2}{b^2} + 4 \\ &= \left(\frac{a^2}{b^2} + 2\right)^2 - \frac{4a^2}{b^2} - \frac{5a^2}{b^2} \\ &= \left(\frac{a^2}{b^2} + 2\right)^2 - \left(\frac{3a}{b}\right)^2 \end{aligned}$$
$\left(\dfrac{a^2}{b^2} + \dfrac{3a}{b} + 2\right)\left(\dfrac{a^2}{b^2} - \dfrac{3a}{b} + 2\right)$
(Each bracket splits again: in full, $\left(\dfrac{a}{b} + 1\right)\left(\dfrac{a}{b} + 2\right)\left(\dfrac{a}{b} - 1\right)\left(\dfrac{a}{b} - 2\right)$.)

$$\begin{aligned} &\frac{p^4}{q^4} + \frac{p^2}{q^2} + 1 \\ &= \left(\frac{p^2}{q^2} + 1\right)^2 - \frac{2p^2}{q^2} + \frac{p^2}{q^2} \\ &= \left(\frac{p^2}{q^2} + 1\right)^2 - \left(\frac{p}{q}\right)^2 \end{aligned}$$
$\left(\dfrac{p^2}{q^2} + \dfrac{p}{q} + 1\right)\left(\dfrac{p^2}{q^2} - \dfrac{p}{q} + 1\right)$

$$\begin{aligned} &\frac{a^4}{b^4} - \frac{7a^2}{b^2} + 1 \\ &= \left(\frac{a^2}{b^2} + 1\right)^2 - \frac{2a^2}{b^2} - \frac{7a^2}{b^2} \\ &= \left(\frac{a^2}{b^2} + 1\right)^2 - \left(\frac{3a}{b}\right)^2 \end{aligned}$$
$\left(\dfrac{a^2}{b^2} + \dfrac{3a}{b} + 1\right)\left(\dfrac{a^2}{b^2} - \dfrac{3a}{b} + 1\right)$

$$\begin{aligned} x^4 + \frac{1}{x^4} - 7 &= \left(x^2 + \frac{1}{x^2}\right)^2 - 2 - 7 \\ &= \left(x^2 + \frac{1}{x^2}\right)^2 - 3^2 \end{aligned}$$
$\left(x^2 + 3 + \dfrac{1}{x^2}\right)\left(x^2 - 3 + \dfrac{1}{x^2}\right)$
(The second bracket splits again: $x^2 + \dfrac{1}{x^2} - 3 = \left(x - \dfrac{1}{x}\right)^2 - 1 = \left(x - \dfrac{1}{x} + 1\right)\left(x - \dfrac{1}{x} - 1\right)$.)

As in Example 6:
$$\begin{aligned} &(p - 5)^2 - 25 + 24 + 6q - 9q^2 \\ &= (p - 5)^2 - (1 - 6q + 9q^2) \\ &= (p - 5)^2 - (1 - 3q)^2 \\ &= (p - 5 + 1 - 3q)(p - 5 - 1 + 3q) \end{aligned}$$
(p − 3q − 4)(p + 3q − 6)