Question

given that a+b+c=5 ab+bc+ca=10 prove that a cube + b cube + c cube - 3ab=-25

Answer 1

Here's your answer!!

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It's given that,

$= > a + b + c = 5........(1)$

And,

$= > ab + bc + ca = 10.....(2)$

We know that,

$= > {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca)$

By substituting value of eq (1) and (2), we get :-

$= > {(5)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(10)$

$= > 25 = {a}^{2} + {b}^{2} + {c}^{2} + 20$

$= > 25 - 20 = {a}^{2} + {b}^{2} + {c}^{2}$

$= > 5 = {a}^{2} + {b}^{2} + {c}^{2} ......(3)$

Now,

We know that,

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

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

By substituting value of equation ,(1) (2)and (3),we get ;-

$= > {a}^{3} + {b}^{3} + {c}^{3} = (5) (5 - 10)$

$= > {a}^{3} + {b}^{3} + {c}^{3} - 3abc = 5 \times ( - 5)$

$= > {a}^{3} + {b}^{3} + {c}^{3} - 3abc = - 25$

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Hope it helps you !! :)