If $c = a^{3} - b^{3},$ then is $c$ a perfect cube?

Yes

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Maybe

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Cannot be determined

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A No
Let $c = x^{3}$ .
Now, $x^{3} = a^{3} - b^{3}$
So, $x^{3} + b^{3} = a^{3}$
Sum of two cubes cannot be another cube.
Ans- Option $B$ .

Suggest Corrections

Similar questions

> 0

\frac{a^{3}}{b^{3}} + \frac{b^{3}}{c^{3}} + \frac{c^{3}}{a^{3}} \geq 3

a + b + c = 0

(a + b - c)^{3} + (b + c - a)^{3} + (c + a - b)^{3}

If $a + b + c = 0$ , then the value of $(a + b - c)^{3} + (b + c - a)^{3} + (c + a - b)^{3}$ is:

a + b + c = 9

(3 - a)^{3} + (3 - b)^{3} + (3 - c)^{3} - 3 (3 - a) (3 - b) (3 - c)

a + b + c = 0,

(a^{3} + b^{3} + c^{3})^{2} =

Join BYJU'S Learning Program