Question

Given a+b+c=6 and ab+bc+ac=11 then find the value of a³+b³+c³-3abc​

Answer 1

To Find :

• we need to find the value of a³+b³+c³-3abc.

Solution :

• a + b + c = 6
• ab + bc + ca = 11
• a³+b³+c³-3abc = ?

We know that,

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

We have value of (a + b + c) and ab + bc + ca so firstly we should find the Value of a² + b² + c².

As we know that,

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

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

›› a² + b² + c² = (6)² - 2 (11)

›› a² + b² + c² = 35 - 22

›› a² + b² + c² = 13

Now,

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

›› a³+ b³+ c³- 3abc = 6[13 -(11)]

›› a³+ b³+ c³- 3abc = 6 (13 - 11)

›› a³+ b³+ c³- 3abc = 6 × 2

›› a³+ b³+ c³- 3abc = 12

Hence,

• Value of a³+ b³+ c³- 3abc is 12 .

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Answer 2

$\red {\bf {\underline {Given:-}}}$

✷ ${\tt{ a + b + c = 6}}$

✷ ${\tt {ab + bc + ca = 11}}$

$\red {\bf {\underline {To\:Find:-}}}$

✷ ${\tt {{a}^{3}+{b}^{3}+{c }^{3} - 3abc}}$

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

We know that,

$\purple{\tt{\star \: {{(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca}}}$

${\tt{⟹(a + b + c)^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca)}}$

${ \tt{⟹ {6}^{2} = ({a}^{2} + {b}^{2} + {c}^{2}) + 2 \times 11}}$

${\tt{⟹36 = ( {a}^{2} + {b}^{2} + {c}^{2}) + 22}}$

${ \tt{⟹36 - 22 = {a}^{2} + {b}^{2} + {c}^{2} }}$

${ \tt{⟹ {a}^{2} + {b}^{2} + {c}^{2} = 14}}$

______________________________

Now, we know that,

$\purple{\star \: {\tt{ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = ( a+ b+c )( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) }}}$

${ \tt{⟹{a}^{3} + {b}^{3} + {c}^{3} - 3abc = ( a+ b+c )( {a}^{2} + {b}^{2} + {c}^{2} -( ab + bc + ca) )}}$

${ \tt{⟹{a}^{3} + {b}^{3} + {c}^{3} - 3abc = 6 \times (14 - 11)}}$

${ \tt{⟹{a}^{3} + {b}^{3} + {c}^{3} - 3abc = 6 \times 3}}$

${ \tt{⟹{a}^{3} + {b}^{3} + {c}^{3} - 3abc = 18}}$

• $\blue {\bf {So,\: the \:value\: of\: {a}^{3} + {b}^{3} + {c}^{3} - 3abc \:is \:18.}}$