Question

If a, b, c are the roots of
x³- x² + 8x – 6 = Othen
(1 +a²)(1+b²)(1+c²) =​

Answer 1

Topic

Complex numbers, polynomials

Hint: Use the identity to first factorize the value to find. We can factorize the given value, by the use of complex numbers.

→ $a ^{2}+1$ is equal to $(a)^{2}-(i)^{2}=(a+i)(a-i)$.

→ $b^{2}+1$ is equal to $(b)^{2}-(i)^{2}=(b+i)(b-i)$.

→ $c^{2}+1$ is equal to $(c)^{2}-(i)^{2}=(c+i)(c-i)$.

Solution

Let a, b, c be the three roots of the equation $x^{3}-x^{2}+8x-6=0$. By factor theorem, $(x-a )(x-b )(x-c )=0$ is equivalent to the equation.

Hence, $x^{3}-x^{2}+8x-6=(x-a )(x-b )(x-c )$ is an identity.

Let's slightly change the equation to avoid confusion.

$x^{3}-x^{2}+8x-6=-(a-x )(b-x )(c-x )$

Substitute $x=i$ into the equation.

$\implies x=i$

$\implies i^{3}-i^{2}+8i-6=-(a-i)(b-i)(c-i)$

$\implies -i-(-1)+8i-6=-(a-i )(b-i )(c-i )$

$\implies -5+7i=-(a-i )(b-i )(c-i )\ \ \ \ \ \bold{...(1)}$

For the next value, let's substitute $x=-i$.

$\implies x=-i$

$\implies (-i)^{3}-(-i)^{2}+8(-i)-6=-(a+i )(b+i )(c+i )$

$\implies i+1-8i-6=-(a+i )(b+i )(c+i )$

$\implies -5-7i=-(a+i )(b+i )(c+i )\ \ \ \ \ \bold{...(2)}$

Multiplying $\bold{(1)}$ and $\bold{(2)}$,

$\implies 5^{2}+7^{2}=(a^{2}+1)(b^{2}+1)(c^{2}+1)$

$\implies 74=(a^{2}+1)(b^{2}+1)(c^{2}+1)$

Hence the required value is $\boxed{74}$.