Question

Without actually calculating the cubes, find the value of (28) ³+(-15) ³+(-13) ³

Answer 1

a³+b³+c³ = (a + b + c) (a² + b² + c² – ab – bc – ca) + 3abc.

a = 28, b = -15 and c = -13

→ = (28)³+(-15)³+(-13)³

→ = (28+(-15)+(-13)) ((28)²+(-15)²+(-13)²-(28)(-15)-(-15)(-13)-(-13)(28)) + 3(28)(-15)(-13)

→ = (28-15-13) ((28)²+(-15)²+(-13)²-(28)(-15)-(-15)(-13)-(-13)(28)) + 3(28)(195)

→ = (28-28)((28)²+(-15)²+(-13)²-(28)(-15)-(-15)(-13)-(-13)(28))+(84)(195)

→ = 0 ((28)²+(-15)²+(-13)²-(28)(-15)-(-15)(-13)-(-13)(28)) + 16380

→ = 0 + 16380

→ = 16380

Hope it helps

Answer 2

Question:

Without actually calculating the cubes, find the value of 28³ + (-15)³ + (13)³.

Answer:

Let a = 28, b = -15 and c = -13, then

a + b + c = 28 - 15 - 13 = 0

__________________

We know that if a + b + c = 0,

then a³ + b³+ c³ = 3abc

__________________

Putting back the values of a, b and c in a³ + b³ + c³ = 3abc

$\therefore$ 28³ + (-15)³ + (-13)³ = 3(28)(-15)(-13)

= 84$\times$195

= 16380.