If x\sqrt{x}x+ 3x3\over\sqrt{x}x3 = 4, then what is the value of x2(x2 – 82)? A-9 B-36 C-81 D36 Show answer & explanationCorrect answer: CExplanationThe correct answer is option 3 i.e. -81 x+3x=4\sqrt{x} + \frac{3}{\sqrt{x}} = 4x+x3=4 Let, y=xy = \sqrt{x}y=x ⇒y+3y=4⇒y2+3=4y⇒y2−4y+3=0\Rightarrow y + \frac{3}{y} = 4 \Rightarrow y^2 + 3 = 4y \Rightarrow y^2 - 4y + 3 = 0 ⇒y+y3=4⇒y2+3=4y⇒y2−4y+3=0 ⇒(y−1)(y−3)=0⇒y=1 or y=3, x=y2⇒x=1 or x=9\Rightarrow (y - 1)(y - 3) = 0 \Rightarrow y = 1 \text{ or } y = 3 \text{, } x = y^2 \Rightarrow x = 1 \text{ or } x = 9⇒(y−1)(y−3)=0⇒y=1 or y=3, x=y2⇒x=1 or x=9 x=1:12(12−82)=1×(−81)=−81x = 1: \quad 1^2(1^2 - 82) = 1 \times (-81) = -81 x=1:12(12−82)=1×(−81)=−81 x=9:92(92−82)=81×(−1)=−81x = 9: \quad 9^2(9^2 - 82) = 81 \times (-1) = -81x=9:92(92−82)=81×(−1)=−81