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Question

# Factorise. (1) x3 + 64y3 (2) 125p3 + q3 (3) 125k3 + 27m3 (4) 2l3 + 432m3 (5) 24a3 + 81b3 (6) y3 + $\frac{1}{8{y}^{3}}$ (7) a3 + $\frac{8}{{a}^{3}}$ (8) 1 + $\frac{{q}^{3}}{125}$

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Solution

## It is known that, ${a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}+{b}^{2}-ab\right)$ $\left(1\right){x}^{3}+64{y}^{3}\phantom{\rule{0ex}{0ex}}={\left(x\right)}^{3}+{\left(4y\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left(x+4y\right)\left\{{\left(x\right)}^{2}+{\left(4y\right)}^{2}-\left(x\right)×\left(4y\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(x+4y\right)\left({x}^{2}+16{y}^{2}-4xy\right)$ $\left(2\right)125{p}^{3}+{q}^{3}\phantom{\rule{0ex}{0ex}}={\left(5p\right)}^{3}+{\left(q\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left(5p+q\right)\left\{{\left(5p\right)}^{2}+{\left(q\right)}^{2}-\left(5p\right)×\left(q\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(5p+q\right)\left(25{p}^{2}+{q}^{2}-5pq\right)$ $\left(3\right)125{k}^{3}+27{m}^{3}\phantom{\rule{0ex}{0ex}}={\left(5k\right)}^{3}+{\left(3m\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left(5k+3m\right)\left\{{\left(5k\right)}^{2}+{\left(3m\right)}^{2}-\left(5k\right)×\left(3m\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(5k+3m\right)\left(25{k}^{2}+9{m}^{2}-15km\right)$ $\left(4\right)2{l}^{3}+432{m}^{3}\phantom{\rule{0ex}{0ex}}=2\left[{l}^{3}+216{m}^{3}\right]\phantom{\rule{0ex}{0ex}}=2\left[{\left(l\right)}^{3}+{\left(6m\right)}^{3}\right]\phantom{\rule{0ex}{0ex}}=2\left[\left(l+6m\right)\left\{{\left(l\right)}^{2}+{\left(6m\right)}^{2}-\left(l\right)×\left(6m\right)\right\}\right]\phantom{\rule{0ex}{0ex}}=2\left(l+6m\right)\left({l}^{2}+36{m}^{2}-6lm\right)$ $\left(5\right)24{a}^{3}+81{b}^{3}\phantom{\rule{0ex}{0ex}}=3\left[8{a}^{3}+27{b}^{3}\right]\phantom{\rule{0ex}{0ex}}=3\left[{\left(2a\right)}^{3}+{\left(3b\right)}^{3}\right]\phantom{\rule{0ex}{0ex}}=3\left[\left(2a+3b\right)\left\{{\left(2a\right)}^{2}+{\left(3b\right)}^{2}-\left(2a\right)×\left(3b\right)\right\}\right]\phantom{\rule{0ex}{0ex}}=3\left(2a+3b\right)\left(4{a}^{2}+9{b}^{2}-6ab\right)$ $\left(6\right){y}^{3}+\frac{1}{8{y}^{3}}\phantom{\rule{0ex}{0ex}}={\left(y\right)}^{3}+{\left(\frac{1}{2y}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left(y+\frac{1}{2y}\right)\left\{{\left(y\right)}^{2}+{\left(\frac{1}{2y}\right)}^{2}-\left(y\right)×\left(\frac{1}{2y}\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(y+\frac{1}{2y}\right)\left({y}^{2}+\frac{1}{4{y}^{2}}-\frac{1}{2}\right)$ $\left(7\right){a}^{3}+\frac{8}{{a}^{3}}\phantom{\rule{0ex}{0ex}}={\left(a\right)}^{3}+{\left(\frac{2}{a}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left(a+\frac{2}{a}\right)\left\{{\left(a\right)}^{2}+{\left(\frac{2}{a}\right)}^{2}-\left(a\right)×\left(\frac{2}{a}\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(a+\frac{2}{a}\right)\left({a}^{2}+\frac{4}{{a}^{2}}-2\right)$ $\left(8\right)1+\frac{{q}^{3}}{125}\phantom{\rule{0ex}{0ex}}={\left(1\right)}^{3}+{\left(\frac{q}{5}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left(1+\frac{q}{5}\right)\left\{{\left(1\right)}^{2}+{\left(\frac{q}{5}\right)}^{2}-\left(1\right)×\left(\frac{q}{5}\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(1+\frac{q}{5}\right)\left(1+\frac{{q}^{2}}{25}-\frac{q}{5}\right)$

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