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Question

# Simplify: (1) (x + y)3 − (x − y)3 (2) (3a + 5b)3 − (3a − 5b)3 (3) (a + b)3 − a3 − b3 (4) p3 − (p + 1)3 (5) (3xy − 2ab)3 − (3xy + 2ab)3

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Solution

## It is known that, ${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+{b}^{2}+ab\right)$ $\left(1\right){\left(x+y\right)}^{3}-{\left(x-y\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left\{\left(x+y\right)-\left(x-y\right)\right\}\left\{{\left(x+y\right)}^{2}+{\left(x-y\right)}^{2}+\left(x+y\right)\left(x-y\right)\right\}\phantom{\rule{0ex}{0ex}}=2y\left({x}^{2}+{y}^{2}+2xy+{x}^{2}+{y}^{2}-2xy+{x}^{2}-{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=2y\left(3{x}^{2}+{y}^{2}\right)$ $\left(2\right){\left(3a+5b\right)}^{3}-{\left(3a-5b\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left\{\left(3a+5b\right)-\left(3a-5b\right)\right\}\left\{{\left(3a+5b\right)}^{2}+{\left(3a-5b\right)}^{2}+\left(3a+5b\right)\left(3a-5b\right)\right\}\phantom{\rule{0ex}{0ex}}=10b\left(9{a}^{2}+25{b}^{2}+30ab+9{a}^{2}+25{b}^{2}-30ab+9{a}^{2}-25{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=10b\left(27{a}^{2}+25{b}^{2}\right)$ $\left(3\right){\left(a+b\right)}^{3}-{a}^{3}-{b}^{3}\phantom{\rule{0ex}{0ex}}={\left(a+b\right)}^{3}-\left({a}^{3}+{b}^{3}\right)\phantom{\rule{0ex}{0ex}}={\left(a+b\right)}^{3}-\left\{\left(a+b\right)\left({a}^{2}+{b}^{2}-ab\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(a+b\right)\left\{{\left(a+b\right)}^{2}-\left({a}^{2}+{b}^{2}-ab\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(a+b\right)\left({a}^{2}+{b}^{2}+2ab-{a}^{2}-{b}^{2}+ab\right)\phantom{\rule{0ex}{0ex}}=\left(a+b\right)\left(3ab\right)\phantom{\rule{0ex}{0ex}}=3{a}^{2}b+3a{b}^{2}\phantom{\rule{0ex}{0ex}}$ $\left(4\right){p}^{3}-{\left(p+1\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left\{p-\left(p+1\right)\right\}\left\{{\left(p\right)}^{2}+{\left(p+1\right)}^{2}+\left(p\right)×\left(p+1\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(-1\right)\left({p}^{2}+{p}^{2}+1+2p+{p}^{2}+p\right)\phantom{\rule{0ex}{0ex}}=\left(-1\right)\left(3{p}^{2}+3p+1\right)\phantom{\rule{0ex}{0ex}}=-3{p}^{2}-3p-1$ $\left(5\right){\left(3xy-2ab\right)}^{3}-{\left(3xy+2ab\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left\{\left(3xy-2ab\right)-\left(3xy+2ab\right)\right\}\left\{{\left(3xy-2ab\right)}^{2}+{\left(3xy+2ab\right)}^{2}+\left(3xy-2ab\right)\left(3xy+2ab\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(-4ab\right)\left(9{x}^{2}{y}^{2}+4{a}^{2}{b}^{2}-12abxy+9{x}^{2}{y}^{2}+4{a}^{2}{b}^{2}+12abxy+9{x}^{2}{y}^{2}-4{a}^{2}{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(-4ab\right)\left(27{x}^{2}{y}^{2}+4{a}^{2}{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=-108{x}^{2}{y}^{2}ab-16{a}^{3}{b}^{3}$

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