Question

The value of parameter a belongs to R such that 5he roots a,b,c of equation x3+6x2+ax-a=0 satisfy the relation (a+3)3+(b+3)3+(c+3)3 is​

Answer 1

Step-by-step explanation:

The value of parameter a belongs to R such that the roots a,b,c of equation

$x^3+6x^2+ax-a=0$

satisfy relation

$(a+3)^3+(b+3)^3+(c+3)^3$

$= ({a}^{3} + 9 + 9a(a + 3)) + ({b}^{3} + 9 + 9b(b + 3)) + ({c}^{3} + 9 + 9c(c + 3))$

$= ( {a}^{3} + {b}^{3} + {c}^{3} ) + 9( {a}^{2} + {b}^{2} + {c}^{2} ) + 27(a + b + c) + 27$

find a.

SOLUTION :-

$f(a) = {a}^{3} + 6 {a}^{2} + {a}^{2} - a \: = 0$

$f(b) = {b}^{3} + 6 {b}^{2} + ab - a = 0$

$f(c) = {c}^{3} + 6 {c}^{2} + ac - a = 0$

Now, add f(a) + f(b) + f(c)

$f(a) + f(b) + f(c) =( {a}^{3} + 6 {a}^{2} + {a}^{2} - a \:) +( {b}^{3} + 6 {b}^{2} + ab - a) + ( {c}^{3} + 6 {c}^{2} + ac - a) = 0$

$( {a}^{3} + {b}^{3} + {c}^{3}) + 6( {a}^{2} + {b}^{2} + {c}^{2} ) + a(a + b + c) - 3a = 0$

So,

$( {a}^{3} + {b}^{3} + {c}^{3}) + 9( {a}^{2} + {b}^{2} + {c}^{2} ) + 27(a + b + c) + 27= 3( {a}^{2} + {b}^{2} + {c}^{2} ) + (27 - a)(a + b + c) + (27 + 3a)$

I THINK YOU HAVE MISSED SOMETHING IN QUESTION TO TYPE IT. SPECIALLY IN RELATION PART.

Hope you get it. Best Of Luck