Question

Help me its very urgent i want the solution

Answer 1

$a + b + c = 9 \\squaring \: both \: sides \\ (a + b + c) {}^{2} = (9 ){}^{2} \\ a {}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) = 81$
$it \: is \: given \: that \: a {}^{2} + {b}^{2} + {c}^{2} = 35$
So,
$35 + 2( ab+ bc+ ca) = 81 \\ 2(ab + bc+ ca) = 81 - 35 \\ ab + bc + a = \frac{46}{2} \\ ab +bc +c a = 23$
Now,
We know that
$a {}^{3} + {b}^{3} + {c}^{3} - 3abc \\ = ( a+ b + c)(a {}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) \\ = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - (ab + bc + ca) \: )$
It is given that ,
$a + b + c = 9(given) \\ a {}^{2} + {b}^{2 } + {c}^{2} = 35 (given)\\ ab + bc + ca = 23 (calculated \: above)$
So,
$= (9) \times (35 - (23) \: ) \\ = 9 \times 12 \\ = 108$
$so \\ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = 108$


----------Hope this will help you-----------

Answer 2

Hey friend ,

Here is your answer ,
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a + b + c = 9
a² + b² + c² = 35

(a+b+c)² = a²+b²+c²+2(ab+bc+ca)

81 = 35 + 2(ab+bc+ca)
46 = 2(ab+bc+ca)
23 = ab+bc+ca

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

a³+b³+c³ - 3abc
= 9(35-23)
= 9*12
= 108                         
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Hope it helped you !!
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