Question

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Without actually finding the cubes find the value of (-23)^3 +(13)^3 +(10)^3 ​

Answer 1

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

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have a great day

Answer 2

Solution:

Given to find:

= (-23)³ + (13)³ + (10)³

We know that if a, b, c are three numbers such that a + b + c = 0, then,

→ a³ + b³ + c³ = 3abc

Here,

→ (-23) + (13) + (10) = 0

So,

(-23)³ + (13)³ + (10)³

= 3 × (-23) × 13 × 10

= -8970 (Answer)

Concept Used:

If a, b and c are 3 numbers such that a + b + c = 0, then,

→ a³ + b³ + c³ = 3abc

Proof:

Given,

→ a + b + c = 0

→ a + b = -c

Cubing both sides, we get,

→ (a + b)³ = -c³

→ a³ + b³ + 3ab(a + b) = -c³

Also, a + b = -c,

→ a³ + b³ + 3ab × (-c) = -c³

→ a³ + b³ - 3abc = -c³

→ a³ + b³ + c³ = 3abc

∆ Using this formula, problem is solved.