Question

If $\alpha ,\beta ,\upsilon$ are roots of cubic $x^3-2x^2+3x+1=0$ then the value of $({\alpha }^3+1)({\beta }^3+1)({\upsilon }^3+1)$ is equal to -

• A. $19$
• B. $23$
• C. $35$
• D. $70$

Answer 1

Answer: (C)

$35$

Sum of roots,$\alpha +\beta +\upsilon =\frac{-(-2)}{1}=2$

and $\alpha \beta +\beta \upsilon +\upsilon \alpha =3$

and $\alpha \beta \upsilon =-1$

$\alpha ,\beta ,\upsilon$ are the roots of the equation, they satisfy equation.

Thus, ${\alpha }^3-2{\alpha }^2+3\alpha +1=0$

${\alpha }^3+1=2{\alpha }^2-3\alpha =\alpha (2\alpha -3)$

Similarly, ${\beta }^3+1=(2\beta -3)\beta$ and ${\upsilon }^3+1=(2\upsilon -3)\upsilon$

Now, $({\alpha }^3+1)({\beta }^3+1)({\upsilon }^3+1)=\alpha (2\alpha -3)\times (2\beta -3)\beta \times (2\upsilon -3)\upsilon$

solving RHS we get,$({\alpha }^3+1)({\beta }^3+1)({\upsilon }^3+1)=\alpha \beta \upsilon [8\alpha \beta \upsilon -12(\alpha \beta +\beta \upsilon +\upsilon \alpha )+18(\alpha +\beta +\upsilon )-27]$

Substituting from above $\Rightarrow (-1)\left[8\right(-1)-12(3)+18(2)-27]$

$({\alpha }^3+1)({\beta }^3+1)({\upsilon }^3+1)=35$