Question

x^3-4x^2-x+1=(x-2)^3​

Answer 1

Given:-

• $\sf {x}^{3} - 4 {x}^{2} - x + 1 = {(x - 2)}^{3}$

To find:-

• The value of x

Answer:-

Given that,

$\sf {x}^{3} - 4 {x}^{2} - x + 1 = {(x - 2)}^{3}$

Expanding RHS using (a - b)³ = a³ - b³ - 3ab(a - b)

$\sf \longrightarrow {x}^{3} - 4 {x}^{2} - x + 1 = {x}^{3} - 2 {}^{3} - 3(x)(2)(x - 2)$

$\sf \longrightarrow {x}^{3} - 4 {x}^{2} - x + 1 = {x}^{3} - 8 - 6x(x - 2)$

$\sf \longrightarrow \cancel{{x}^{3}} - 4 {x}^{2} - x + 1 = \cancel{{x}^{3}} - 8 - 6 {x}^{2} + 12x$

$\sf \longrightarrow 6 {x}^{2} - 4 {x}^{2} - x - 12x + 1 + 8 = 0$

$\sf \longrightarrow 2 {x}^{2} - 13x + 9 = 0$

On comparing with $\sf ax^2 + bx + c = 0,$ we get,

• a = 2
• b = -13
• c = 9

By quadratic formula,

$\sf \longrightarrow x_{1,2} = \dfrac{ - b \pm \sqrt{b {}^{2} - 4ac } }{2a}$

$\sf \longrightarrow x_{1,2} = \dfrac{ - ( - 13) \pm \sqrt{ {( - 13)}^{2} - (4 \times 2 \times 9) } }{2 \times 2}$

$\sf \longrightarrow x_{1,2} = \dfrac{ 13 \pm \sqrt{ {169 - 72 } }}{2 \times 2}$

$\sf \longrightarrow x_{1,2} = \dfrac{ 13 \pm \sqrt{ {97}}}{4}$

Hence, the two solutions are,

$\longrightarrow \underline{ \underline{ \sf{ \green{ x_{1} = \dfrac{13 + \sqrt{97} }{4} }}}}$

$\longrightarrow \underline{ \underline{ \sf{ \green{ x_{2} = \dfrac{13 - \sqrt{97} }{4} }}}}$