Question

If a + b + c = 15 and a2 + b2 + c2 = 83, find the value of a3 + b3 + c3 – 3abc.

Answer 1

Solution

Given :-

• a + b + c = 15 -----------(1)
• a² + b² + c² = 83 --------(2)

Find :-

• a³ + b³ + c³ - 3abc = ?

Explanation

Using Formula

★ ( x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

★ x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - 3xyz)

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Squaring both side of equ(1)

==> (a + b + c)² = 15²

==> a² + b² + c² + 2(ab + bc + ca) = 225

Keep Value by equ(2)

==> 83 + 2(ab + bc + ca) = 225

==> 2(ab + bc + ca) = 225 - 83

==> 2(ab + bc + ca) = 142

==> (ab + bc + ca) = 142/2

==> (ab + bc + ca) = 71 -------------(3)

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Now, Calculate

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

Keep Value by equ(1) , equ(2) & equ(3)

==> (a³ + b³ + c³ - 3abc) = (15)(83 - 71)

==> (a³ + b³ + c³ - 3abc) = 15 * 12

==> (a³ + b³ + c³ - 3abc) = 180 .

[ Ans ]

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Answer 2

Given -

In the above Question , we have the following information given -

$\sf{ a + b + c = 15 \: and \: a^2 + b^2 + c^2 = 83 }$

We have to find the value of $\sf{ a^3 + b^3 + c^3 - 3abc }$

Solution -

$\sf{ a + b + c = 15 } \\ \\ \sf{ Squaring \ both \ sides \ - } \\ \\ \sf{ {( a + b + c )} ^ 2 = { 15} ^ 2 = 225 } \\ \\ \sf{ => a^2 + b^2 + c^2:+ 2ab + 2bc + 2ac = 225 } \\ \\ \sf{ But \: a^2 + b^2 + c^2 = 83 } \\ \\ \sf{ \bold{ Hence \: - }} \\ \\ \sf{ => 83 + 2ab + 2bc + 2ac = 225 } \\ \\ \sf{ => 2ab + 2bc + 2ac = 225 - 83 = 142 } \\ \\ \sf{ => ab + bc + ac = 71 \: ..... ....... \: [1] } \\ \\ \sf{ a^3 + b^3 + c^3 - 3abc } \\ \\ \sf{ => ( a + b + c ) \times ( a^2 + b^2 + c^2 - ab - bc - ac ) } \\ \\ \sf{ \bold{ Substituting \: the \: given \: Values \: - }} \\ \\ \sf{ => ( 15 ) \times ( 83 71 ) } \\ \\ \sf{=> 15 \times 12 } \\\\ \sf{ = > 180 \: ........... [A] }$