Question

Without actually calculating the cubes, find the value of the following:

${\left(28\right)}^3+{(-15)}^3+{(-13)}^3$

Answer 1

Find the value of ${\left(28\right)}^3+{(-15)}^3+{(-13)}^3$

We know, ${\mathbf{a}}^{\mathbf{3}}\mathbf{+}{\mathbf{b}}^{\mathbf{3}}\mathbf{+}{\mathbf{c}}^{\mathbf{3}}\mathbf{-}\mathbf{3}\mathbf{abc}\mathbf{=}\left(a+b+c\right)\left(a^2+b^2+c^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca}\right)$

And, ${\mathbf{a}}^{\mathbf{3}}\mathbf{+}{\mathbf{b}}^{\mathbf{3}}\mathbf{+}{\mathbf{c}}^{\mathbf{3}}\mathbf{=}\mathbf{3}\mathbf{abc}$ if $\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}\mathbf{=}\mathbf{0}$.

Now, here $a=28,b=-15,c=-13$

And sum is

$$\begin{gathered} =28+\left(-15\right)+\left(-13\right) \\ =28-15+-13 \\ =0 \end{gathered}$$

Now,

$$\begin{gathered} {\left(28\right)}^3+{(-15)}^3+{(-13)}^3=3\times 28\times -15\times -13 \\ =16380 \end{gathered}$$

Thus, value of ${\left(28\right)}^3+{(-15)}^3+{(-13)}^3=16380$