Question

Expand
(1) (2m − 5)³
(2) (4 − p)³
(3) (7x − 9y)³
(4) (58)³
(5) (198)³
(6) ${\left(2p-\frac{1}{2p}\right)}^3$
(7) ${\left(1-\frac{1}{a}\right)}^3$
(8) ${\left(\frac{x}{3}-\frac{3}{x}\right)}^3$

Answer 1

It is known that,
${\left(a-b\right)}^3=a^3-b^3-3a^{2}b+3a{b}^2$
$$\begin{gathered} \left(1\right){\left(2m-5\right)}^3 \\ ={\left(2m\right)}^3-{\left(5\right)}^3-3\times {\left(2m\right)}^2\times 5+3\times 2m\times {\left(5\right)}^2 \\ =8m^3-125-60m^2+150m \end{gathered}$$
$$\begin{gathered} \left(2\right){\left(4-p\right)}^3 \\ ={\left(4\right)}^3-{\left(p\right)}^3-3\times {\left(4\right)}^2\times \left(p\right)+3\times 4\times {\left(p\right)}^2 \\ =64-p^3-48p+12p^2 \end{gathered}$$
$$\begin{gathered} \left(3\right){\left(7x-9y\right)}^3 \\ ={\left(7x\right)}^3-{\left(9y\right)}^3-3\times {\left(7x\right)}^2\times 9y+3\times 7x\times {\left(9y\right)}^2 \\ =343x^3-729y^2-1323x^{2}y+1701x{y}^2 \end{gathered}$$
$$\begin{gathered} \left(4\right){\left(58\right)}^3 \\ ={\left(60-2\right)}^3 \\ ={\left(60\right)}^3-{\left(2\right)}^3-3\times {\left(60\right)}^2\times \left(2\right)+3\times 60\times {\left(2\right)}^2 \\ =216000-8-21600+720 \\ =195112 \end{gathered}$$
$$\begin{gathered} \left(5\right){\left(198\right)}^3 \\ ={\left(200-2\right)}^3 \\ ={\left(200\right)}^3-{\left(2\right)}^3-3\times {\left(200\right)}^2\times 2+3\times 200\times {\left(3\right)}^2 \\ =8000000-8-240000+5400 \\ =7765392 \end{gathered}$$
$$\begin{gathered} \left(6\right){\left(2p-\frac{1}{2p}\right)}^3 \\ ={\left(2p\right)}^3-{\left(\frac{1}{2p}\right)}^3-3\times {\left(2p\right)}^2\times \frac{1}{2p}+3\times 2p\times {\left(\frac{1}{2p}\right)}^2 \\ =8p^3-\frac{1}{8p^3}-6p+\frac{3}{2p} \end{gathered}$$
$$\begin{gathered} \left(7\right){\left(1-\frac{1}{a}\right)}^3 \\ ={\left(1\right)}^3-{\left(\frac{1}{a}\right)}^3-3\times {\left(1\right)}^2\times \frac{1}{a}+3\times 1\times {\left(\frac{1}{a}\right)}^2 \\ =1-\frac{1}{a^3}-\frac{3}{a}+\frac{3}{a^2} \end{gathered}$$
$$\begin{gathered} \left(8\right){\left(\frac{x}{3}-\frac{3}{x}\right)}^3 \\ ={\left(\frac{x}{3}\right)}^3-{\left(\frac{3}{x}\right)}^3-3\times {\left(\frac{x}{3}\right)}^2\times \left(\frac{3}{x}\right)+3\times \left(\frac{x}{3}\right)\times {\left(\frac{3}{x}\right)}^2 \\ =\frac{x^3}{27}-\frac{27}{x^3}-x+\frac{9}{x} \end{gathered}$$