Question

Is (8/15)3-(1/3)3-(1/5)3=(8/75) ?How will you justify your answer ,without actually calculating the cubes ?

Answer 1

Let a=(8/15), b=(-1/3) and c=(-1/5).
If a+b+c=0, then a3+b3+c3=3abc
here (8/15)+(-1/3)+(-1/5)=0
Hence, without calculating the cubes, the answer is 3*(8/15)*(-1/3)*(-1/5)=8/75.

Answer 2

Answer:

Yes $(\frac{8}{15})^3-(\frac{1}{3})^3 -(\frac{1}{5})^3=\frac{8}{75}$

Step-by-step explanation:

To find : Is $(\frac{8}{15})^3-(\frac{1}{3})^3 -(\frac{1}{5})^3=\frac{8}{75}$ How will you justify your answer ,without actually calculating the cubes ?

Solution :

Let $a=\frac{8}{15}$, $b=-\frac{1}{3}$ and $c=-\frac{1}{5}$

So we have $a^3+b^3+c^3$

Now applying identity,

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$

$a+b+c=\frac{8}{15}+(-\frac{1}{3})+(-\frac{1}{5})$

$a+b+c=\frac{8}{15}-\frac{5+3}{15}$

$a+b+c=\frac{8}{15}-\frac{8}{15}$

$a+b+c=0$

So, $a^3+b^3+c^3-3abc=0$

$a^3+b^3+c^3=3abc$

$(\frac{8}{15})^3+(-\frac{1}{3})^3+(-\frac{1}{5})^3=3\times \frac{8}{15}\times (-\frac{1}{3})\times (-\frac{1}{5})$

$(\frac{8}{15})^3-(\frac{1}{3})^3 -(\frac{1}{5})^3=\frac{8}{75}$

Therefore, Yes $(\frac{8}{15})^3-(\frac{1}{3})^3 -(\frac{1}{5})^3=\frac{8}{75}$