Question

Without actually calculating the cubes, find the value of each 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, ${\left(28\right)}^3+{(-15)}^3+{(-13)}^{3^{}}.$

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

And sum is

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

So,

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

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