Question

,find value of: (-1) cube+ (-2) cube+(-3) cube+(-4) cube+2(5) cube.

Answer 1

$$\begin{gathered} {\left(-1\right)}^3+{\left(-2\right)}^3+{\left(-3\right)}^3+{\left(-4\right)}^3+2{\left(5\right)}^3 \\ ={\left(-1\right)}^3+{\left(-2\right)}^3+{\left(-3\right)}^3+{\left(-4\right)}^3+{\left(5\right)}^3+{\left(5\right)}^3 \\ =\left\{{\left(-1\right)}^3+{\left(-4\right)}^3+{\left(5\right)}^3\right\}+\left\{{\left(-2\right)}^3+{\left(-3\right)}^3+{\left(5\right)}^3\right\} \\ \mathrm{Taking}{\left(-1\right)}^3+{\left(-4\right)}^3+{\left(5\right)}^3 \\ \mathrm{Let}a=-1,b=-4\mathrm{and}c=5 \\ a+b+c=\left(-1\right)+\left(-4\right)+5=0 \\ \mathrm{Since},a+b+c=0 \\ \mathrm{then},a^3+b^3+c^3=3abc \\ \Rightarrow {\left(-1\right)}^3+{\left(-4\right)}^3+{\left(5\right)}^3=3\left(-1\right)\left(-4\right)\left(5\right)=60 \\ \mathrm{Taking}{\left(-2\right)}^3+{\left(-3\right)}^3+{\left(5\right)}^3 \\ \mathrm{Let}x=-2,y=-3\mathrm{and}z=5 \\ x+y+z=\left(-2\right)+\left(-3\right)+5=0 \\ \mathrm{Since},x+y+z=0 \\ \mathrm{then},x^3+y^3+z^3=3xyz \\ \Rightarrow {\left(-2\right)}^3+{\left(-3\right)}^3+{\left(5\right)}^3=3\left(-2\right)\left(-3\right)\left(5\right)=90 \\ \mathrm{Therefore},\left\{{\left(-1\right)}^3+{\left(-4\right)}^3+{\left(5\right)}^3\right\}+\left\{{\left(-2\right)}^3+{\left(-3\right)}^3+{\left(5\right)}^3\right\}=60+90 \\ \Rightarrow {\left(-1\right)}^3+{\left(-2\right)}^3+{\left(-3\right)}^3+{\left(-4\right)}^3+2{\left(5\right)}^3=150 \end{gathered}$$