Question

$<i><b>Good afternoon friends$

$<i><b>Answer the above question$

$<i><b>No Spam$

Answer 1

Given,$x= \bigg( -\dfrac{a}{2} + \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} } \bigg)^{1/3} + \bigg( -\dfrac{a}{2} - \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} } \bigg)^{1/3}$

Cube on both sides,

$x^3 =\bigg[\bigg( -\dfrac{a}{2} + \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} } \bigg)^{1/3} + \bigg( -\dfrac{a}{2} - \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} } \bigg)^{1/3} \bigg]^{3}$

We know, ( a + b )^3 = a^3 + b^3 + 3ab( a + b ).

$\therefore x^3 = \bigg( -\dfrac{a}{2} + \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} } \bigg)^{\frac{1}{3}\times 3} + \bigg( -\dfrac{a}{2} - \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} } \bigg)^{\frac{1}{3} \times 3} + 3\bigg[ \bigg( -\dfrac{a}{2} + \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} } \bigg)^{1/3} + \bigg( -\dfrac{a}{2} - \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} } \bigg)^{1/3} \bigg] \bigg( -\dfrac{a}{2} + \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} } \bigg)^{\frac{1}{3}} \bigg( -\dfrac{a}{2} - \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} } \bigg)^{\frac{1}{3} }$

$\text{In the question, value of}\bigg(-\dfrac{a}{2} + \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} } \bigg)^{1/3} + \bigg( -\dfrac{a}{2} - \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} } \bigg)^{1/3}\text{is given and it is x}$

$\therefore x^3 =-\dfrac{a}{2} + \sqrt{\dfrac{a^2}{4} +\dfrac{b^{3}}{27}}-\dfrac{a}{2} - \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} } + 3x \bigg( -\dfrac{a}{2} + \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} }\bigg)^{} \bigg( -\dfrac{a}{2} - \sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} } \bigg)$

We know, ( a - b )( a + b ) = a^2 - b^2

$\therefore x^3 = -\dfrac{a}{2} - \dfrac{a}{2} + 3x\bigg[ \bigg( -\dfrac{a}{2} \bigg)^2 -\bigg(\sqrt{\dfrac{a^2}{4} + \dfrac{b^3}{ 27} }\bigg)^2 \bigg]$

$x^3 = - a + 3x\bigg(\dfrac{a^2 }{4} - \dfrac{a^2}{4} - \dfrac{b^3}{27} \bigg)^{1/3}\\ \\\\x^3 = - a + 3x\bigg( -\dfrac{b^3}{3^3}\bigg)^{1/3} \\\\ \\ x^3 = - a + 3x \bigg( \dfrac{-b}{3}\bigg)$

$x^3 = - a + (x \times - b)\\\\ x^3 = - a - bx \\\\x^3 + a + bx =0 \\\\ x^3 + bx + a =0$

Therefore, the value of x^3 + bx + a = 0

Option C is correct.

Answer 2

Answer:

hey mate 0 is the correct answer