Question

Without actually calculating the cubes, find the value of 27³+(-14)³+(-13)³

Answer 1

Answer:

Step-by-step explanation:

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

$it \: is \: in \: the \: form \: of \: a {}^{3} + b {}^{3} + c {}^{3} \\ \\ we \: know \: that \: if \: a + b + c = 0 \\ \\ then \: a {}^{3} + b {}^{3} + c {}^{3} = 3abc \\ \\ here \: \: a + b + c \: is \: 0 \: \\ \\ - > 27 {}^{3} + ( - 14) {}^{3} + ( - 13) {}^{3} \\ \\ = 3(27)( - 14)( - 13) = 14742$