Question

if root of the equation of 4x^3-3x^2-6x-4 is in the form of (3√a+3√b+c)/4 then find a+b+c

Answer 1

Given:

The equation is

$4x^{3} -3x^{2}-6x-4$

The form of

$\dfrac{(\sqrt[3]{a}+\sqrt[3]{b}+c) }{4}$

To Find:

The value of $a+b+c$

Step-by-step explanation:

The equation can be written as

$x^{3}=\dfrac{3x^{2}+6x+4 }{4}\\=\dfrac{3}{4}x^{2}+\dfrac{3}{2}x+1$

Assume $x=2u\Rightarrow 8u^{3}=3u^{2}+3u+1$

$9u^{3}=u^{3}+3u^{2}+3u+1=(u+1)^{3}\\=(9^{\frac{1}{3} }u)^{3}\\\Rightarrow 9^{\frac{1}{3}u } =u+1\\(9^{\frac{1}{3} }-1)u=1\\\Rightarrow x=\frac{2}{9^({\frac{1}{3} }-1) }$

$x=\frac{2}{9^{\frac{1}{3}}-1}\left(\frac{9^{\frac{2}{3} }+9^{\frac{1}{3} }+1 }{9^{\frac{2}{3}}+9^{\frac{1}{3} }+1 } \right) \\=\frac{2}{9-1}\left({9^{\frac{2}{3} }+9^{\frac{1}{3} }+1 } \right)\\=\frac{1}{4}\left({9^{\frac{2}{3} }+9^{\frac{1}{3} }+1 } \right)\\=\dfrac{\sqrt[3]{81}+\sqrt[3]{9}+1 }{4} \\a=81, b=9\ and\ c=1$

Therefore the value of

$(a+b+c)=(81+9+1)\\=91.$