Question

If a+b+c=16 and a^2+b^2+c^2=90 then find the value of a^3+b^3+c^3-3abc​

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 + 2ab + 2bc + 2ca

→ 256 - 90 = 2(ab + bc + ca)

→ 166 = 2(ab + bc + ca)

→ 83 = ab + bc + ca

→ 83 = -1(-ab - bc - ca)

→ -83 = -ab - bc - ca

Now,

• 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

$\large{\boxed{\underline{\underline{\bf{\red{Answer:-}}}}}}$

$\implies$ 112

$\large\underline\mathrm{hence,}$

$\large\underline\mathrm{The \: value \: of \: a² \: + \: b² \: + \: c² \: -3abc \: is \: 112 \: .}$

$\large\underline\mathrm{Given:-}$

• a + b + c = 16
• a² + b² + c² = 90

$\large\underline\mathrm{To \: find}$

• value of a² + b² + c² -3abc

$\large\underline\mathrm{So,}$

$\implies$ (a + b + c)² = a² + b² + c² + 3ab + 2bc + 2ca

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

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

$\implies$ 166 = 2(ab + bc + ca)

$\implies$ 83 = (ab + bc + ca)

$\implies$ 83 = -1(-ab - bc - ca)

$\implies$ -83 = -ab - bc - ca

$\large\underline\mathrm{Then,}$

$\implies$ a² + b² + c² -3abc = (a + b + c)(a² + b² + c² - ab - BC - ca)

$\implies$ (16)(90 - 83)

$\implies$ (16)(7)

$\implies$ 112

$\large\underline\mathrm{hence,}$

$\large\underline\mathrm{The \: value \: of \: a² \: + \: b² \: + \: c² \: -3abc \: is \: 112 \: .}$

$\large\underline\mathrm{Hope \: it \: helps \: you \: plz \: mark \: me \: brainlist}$