Question

If a + b + c is equals to 16 and a square + b square + c square is equal to 90 then find the value of a cube plus b cube plus c cube - 3abc

Answer 1

Step-by-step explanation:

Given -

a + b + c = 16

a² + b² + c² = 90

To Find -

• Value of a³ + b³ + c³ - 3abc

As we know that :-

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

→ (16)² = 90 + 2(ab + bc + ca)

→ 256 - 90 = -2(-ab - bc - ca)

→ 166 = -2(-ab - bc - ca)

→ -83 = -ab - bc - ca

Now,

As we know that :-

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

→ (16)(90 - 83)

→ 16 × 7

→ 112

Hence,

The value of a³ + b³ + c³ - 3abc is 112.

Answer 2

$\large{\underline{\bf{\green{Given:-}}}}$

✰ a + b + c = 16

✰ a²+ b² + c² = 90

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰ we need to find the Value of a³ + b³ + c³ - 3abc.

$\huge{\underline{\bf{\red{Solution:-}}}}$

we know that

(a +b+c)²

(a +b+c)² = a² + b²+ c²+ 2ab + 2bc + 2ca

$: \implies \sf\:$(16)² = 90 +2(ab +bc + ca)

$: \implies \sf\:$256 - 90 = -2(- ab - bc - ca)

$: \implies \sf\:$166 = -2(- ab - bc - ca)

$: \implies \sf\:$ 166/-2 = (- ab - bc - ca)

$: \implies \sf\:$ -83 = (- ab - bc - ca)

Now,

we know that:-

a³ + b³ + c³ - 3abc

= ( a + b + c) (a² + b² + c²- ab - bc - ca )

$: \implies \sf\:$16 × (90-83)

$: \implies \sf\:$16 × 7

$: \implies \sf\:$ 112.

So,

a³ + b³ + c³ - 3abc = 112

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