Question

without calculating cubes find the value of (-1)cube + (-2)cube + (-3) cube+ (-4)cube + 2(5) cube

Answer 1

$$\begin{gathered} {\left(-1\right)}^3+{\left(-2\right)}^3+{\left(-3\right)}^3+{\left(-4\right)}^3+2{\left(5\right)}^3 \\ =\left[{\left(-1\right)}^3+{\left(-4\right)}^3+{\left(5\right)}^3\right]+\left[{\left(-2\right)}^3+{\left(-3\right)}^3+{\left(5\right)}^3\right] \\ \mathrm{Consider}\left[{\left(-1\right)}^3+{\left(-4\right)}^3+{\left(5\right)}^3\right] \\ \mathrm{Let}\mathrm{a}=-1;\mathrm{b}=-4;\mathrm{c}=5 \\ \mathrm{Now},\mathrm{a}+\mathrm{b}+\mathrm{c}=-1-4+5=0 \\ \mathrm{So},{\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3=3\mathrm{abc} \\ \Rightarrow {\left(-1\right)}^3+{\left(-4\right)}^3+{\left(5\right)}^3=3\left(-1\right)\left(-4\right)\left(5\right)=60 \\ \mathrm{Consider}\left[{\left(-2\right)}^3+{\left(-3\right)}^3+{\left(5\right)}^3\right] \\ \mathrm{Let}\mathrm{a}=-2;\mathrm{b}=-3;\mathrm{c}=5 \\ \mathrm{Now},\mathrm{a}+\mathrm{b}+\mathrm{c}=-2-3+5=0 \\ \mathrm{So},{\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3=3\mathrm{abc} \\ \Rightarrow {\left(-2\right)}^3+{\left(-3\right)}^3+{\left(5\right)}^3=3\left(-2\right)\left(-3\right)\left(5\right)=90 \\ \mathrm{So},{\left(-1\right)}^3+{\left(-4\right)}^3+{\left(5\right)}^3+{\left(-2\right)}^3+{\left(-3\right)}^3+{\left(5\right)}^3=60+90 \\ \Rightarrow {\left(-1\right)}^3+{\left(-2\right)}^3+{\left(-3\right)}^3+{\left(-4\right)}^3+2{\left(5\right)}^3=150 \end{gathered}$$