Question

x3-4x2-x+1=(x-2)3 ansk fast

Answer 1

Answer :

${x}^{3} - 4 {x}^{2} - x + 1 = {(x - 2)}^{3} \\ \\ = > {x}^{3 } - 4 {x}^{2} - x + 1 = {x}^{3} - 3 {(x)}^{2}(2) + 3(x) {(2)}^{2} - {(2)}^{3} \\ \\ = > {x}^{3} - 4 {x}^{2} - x + 1 = {x}^{3} - 6 {x}^{2} + 12x - 8 \\ \\ = > 2 {x}^{2} - 13x + 9 = 0$

Now, By Shridharacharya Formula :

Here a = 2, b = -13 and c = 9

$x = \frac{ - {b} \binom{ + }{ - } \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ = > x = \frac{ - ( - 13) \binom{ + }{ - } \sqrt{ { ( - 13)}^{2} - 4(2)(9) } }{2 \times 2} \\ \\ = > x = \frac{13 \binom{ + }{ - } \sqrt{169 - 72} }{4} \\ \\ = > x = \frac{13 \binom{ + }{ - } \sqrt{97} }{4} \\ \\ = > x = \frac{13 + \sqrt{97} }{4} ,\: \frac{13 - \sqrt{97} }{4}$

Answer 2

Given:

$x^3-4x^2-x+1=(x-2)^3$

To Find:

Find value of x

Solution:

$x^3-4x^2-x+1=(x-2)^3$

Formula:

$(a-b)^3=a^3-b^3-3a^2b+3ab^2$

Using formula :

$\Rightarrow x^3-4x^2-x+1=x^3-8-3(x)^2(2)+3(x)(2)^2\\ \Rightarrow x^3-4x^2-x+1=x^3-8-6x^2+12x\\ \Rightarrow -4x^2-x+1=-8-6x^2+12x\\ \Rightarrow-4x^2+6x^2+1+8=12x+x\\ \Rightarrow 2x^2+9=13x\\ \Rightarrow 2x^2+6x+3x+9=0\\ \Rightarrow 2x(x+3)+3(x+3)=0\\ \Rightarrow (2x+3)(x+3)=0\\ \Rightarrow x=-3,\frac{-3}{2}$

Hence The value of x is -3 and$-\frac{3}{2}$