Question

Use the formula to find the factors.

(1) 27m³ + 125n³
(2) p³ + 64q³
(3) y³ + 216
(4) 2a³ + 250b³
(5) 343 + 512k³
(6) $a^3+\frac{1}{8a^3}$

Answer 1

(1) 27m³ + 125n³ can be factorised as:

$$\begin{gathered} 27m^3+125n^3 \\ ={\left(3m\right)}^3+{\left(5n\right)}^3 \\ =\left(3m+5n\right)\left[{\left(3m\right)}^2-3m\times 5n+{\left(5n\right)}^2\right]\left[\mathrm{using}\mathrm{the}\mathrm{formula}{a}^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\right] \\ =\left(3m+5n\right)\left(9m^2-15mn+25n^2\right) \end{gathered}$$

(2) p³ + 64q³ can be factorised as:

$$\begin{gathered} p^3+64q^3 \\ =p^3+{\left(4q\right)}^3 \\ =\left(p+4q\right)\left[p^2-p\times 4q+{\left(4q\right)}^2\right]\left[\mathrm{using}\mathrm{the}\mathrm{formula}{a}^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\right] \\ =\left(p+4q\right)\left(p^2-4pq+16q^2\right) \end{gathered}$$

(3) y³ + 216 can be factorised as:

$$\begin{gathered} y^3+216 \\ ={\left(y\right)}^3+{\left(6\right)}^3 \\ =\left(y+6\right)\left(y^2-y\times 6+6^2\right)\left[\mathrm{using}\mathrm{the}\mathrm{formula}{a}^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\right] \\ =\left(y+6\right)\left(y^2-6y+36\right) \end{gathered}$$


(4) 2a³ + 250b³ can be factorised as:

$$\begin{gathered} 2a^3+250b^3 \\ =2\left[a^3+125b^3\right] \\ =2\left[{\left(a\right)}^3+{\left(5b\right)}^3\right] \\ =2\left(a+5b\right)\left[a^2-a\times 5b+{\left(5b\right)}^2\right]\left[\mathrm{using}\mathrm{the}\mathrm{formula}{a}^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\right] \\ =2\left(a+5b\right)\left(a^2-5ab+25b^2\right) \end{gathered}$$

(5) 343 + 512k³ can be factorised as:

$$\begin{gathered} 343+512k^3 \\ ={\left(7\right)}^3+{\left(8k\right)}^3 \\ =\left(7+8k\right)\left[7^2-7\times 8k+{\left(8k\right)}^2\right]\left[\mathrm{using}\mathrm{the}\mathrm{formula}{a}^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\right] \\ =\left(7+8k\right)\left(49-56k+64k^2\right) \end{gathered}$$

(6) $a^3+\frac{1}{8a^3}$ can be factorised as:

$$\begin{gathered} a^3+\frac{1}{8a^3} \\ ={\left(a\right)}^3+{\left(\frac{1}{2a}\right)}^3 \\ =\left(a+\frac{1}{2a}\right)\left[a^2-a\times \frac{1}{2a}+{\left(\frac{1}{2a}\right)}^2\right]\left[\mathrm{using}\mathrm{the}\mathrm{formula}{a}^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\right] \\ =\left(a+\frac{1}{2a}\right)\left(a^2-\frac{1}{2}+\frac{1}{4a^2}\right) \end{gathered}$$