Question

95. Value of a3 + b3 + c3 +3abc if a + b + c = 12 and ab + bc + ca = 47

Answer 1

GIVEN :

The values are a + b + c = 12 and ab + bc + ca = 47

TO FIND :

The value of $a^3 + b^3+c^3- 3abc$

SOLUTION :

Given that the values are a + b + c = 12 and ab + bc + ca = 47

We know the Algebraic identity

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

Now we have to find the value of  $a^2 + b^2 + c^2$ :

By using the Algebraic identity

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

Putting the value of a + b + c = 12 and ab + bc + ca = 47 in the above formula,

$( 12 )^2 = a^2 + b^2 + c^2 + 2 ( 47 )$

$144=a^2+b^2+c^2+94$

$a^2 + b^2 + c^2 = 144-94$

$a^2 + b^2 + c^2=50$

Now, substituting value in the formula for $a^3 + b^3+c^3- 3abc$

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

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

$a^3 + b^3+c^3- 3abc= ( 12 ) ( 50 - (47 ))$

$a^3 + b^3+c^3- 3abc=( 12 ) ( 3 )$

∴ $a^3 + b^3+c^3- 3abc = 36$

∴ the value of $a^3 + b^3+c^3- 3abc$ is 36