Question

(a+b+c)=8 and a2+b2 +c2=26 then find the value of a3+b3 +c3 _3abc

Answer 1

Hi friend ✋✋✋✋
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your answer
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Given that : -

(a + b + c) = 8 , (a² + b² + c²) = 26

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

=> (8)² = 26 + 2(ab + bc + ca)

=> 64 = 26 - 2(ab + bc + ca)

=> 64 - 26 = 2(ab + bc + ca)

=> (ab + bc + ca) = 38/2

=> (ab + bc + ca) = 19

Again,
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a³ + b³ + c³ - 3abc

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

=> (a + b + c) [(a² + b² + c²) - (ab + bc + ca)]

=> 8 × (26 - 19)

=> 8 × 7

=> 56

Therefore,
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a³ + b³ + c³ - 3abc = 56

HOPE IT HELPS

Answer 2

Hey friend , Harish here.

Here is your answer.

$a+b+c=8$ -(i)

$(a ^{2} +b^{2} +c^{2}) = 26$  -(ii)

Now, Square on both side in equation (i).We get,

=>$(a+b+c)^{2} = 8 ^{2}$

=>$(a ^{2} +b^{2} +c^{2}) +2(ab + bc + ca) = 64$  -(iii)

Now substitute value of equation (ii) in eq (iii)

=>$26+2(ab + bc + ca) = 64$

=>$2(ab + bc + ca) = 64 - 26$

=>$2(ab + bc + ca) = 38$

=>$(ab + bc + ca) = 19$

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

Now by applying the above identity we get,

$a ^{3} + b ^{3} + c ^{3} -3abc = (8) ( 26 - 19)$

                                              $= (8) (7)$   

                                              $= 56$
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Hope my answer is helpful to u.