Question

If a,b,c,d are roots of $x^4-5x-3=0$ then value of $a^3+b^3+c^3+d^3$ is

• A. $-10$
• B. $10$
• C. $15$
• D. None of these

Answer 1

The correct option is C $15$

$(x-a)(x-b)(x-c)(x-d)=x^4-5x-3$

$(x^2-(a+b)x+ab)(x^2-(c+d)x+cd)=x^4-5x-3$

$x^4-(a+b)x^3+ab{x}^2-(c+d)x^3+(a+b)(c+d)x^2-ab(c+d)x+cd{x}^2-(a+b)cdx+abcd$

$=x^4-5x-3$

Comparing coefficient :

$x^2:(ab+(a+b\left)\right(c+d)+cd)=0$

$x^3:-(a+b)-(c+d)=0$

$x:-ab(c+d)+(a+b)cd=5$

$x^0:abcd=-3$

Substituting $(a+b)-(c+d)$ from $x^3$ equation is x eqn.

$(ab-cd)(c+d)=5$

$\Rightarrow ab-cd=5(c+d)\to \left(1\right)$

$ab-(c+d{)}^2+cd=0\to (2)$ in $x2$ eqn

$\Rightarrow (ab+cd)=(c+d{)}^2$

$abcd=-3$

$\Rightarrow ab=\frac{-3}{cd}$

Required $a^3+b^3+c^3+d^3$

$=(a+b)(a^2+b^2-ab)+(c+d)(c^2+d^2-cd)$

$=(c+d)(c^2+d^2-cd-a^2-b^2+ab)$

$=(c+d)(c+d{)}^2-3cd-(a+b{)}^2+3ab)$

$=(c+d)(ab-cd)$

$=3(c+d)\frac{5}{(c+d)}=15$

Hence answer is (D)