If $x(x-3)=-1$ then the value of $x^{3}\left(x^{3}-18\right)$ will be,

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Algebra 2

Evaluate expression when two complex numbers are given

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Q. If

x(x-3)=-1

then the value of

x^{3}\left(x^{3}-18\right)

will be,

Given Equation: We are given the equation $x(x-3) = -1$ . $\newline$ To find the value of $x^{3}(x^{3}-18)$ , we first need to find the value of $x$ . $\newline$ Let's solve the given equation for $x$ . $\newline$ $x^2 - 3x + 1 = 0$ $\newline$ This is a quadratic equation, and we can solve it using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = -3$ , and $c = 1$ .
Quadratic Formula: Calculate the discriminant $b^2 - 4ac$ of the quadratic equation. $\newline$ Discriminant = $(-3)^2 - 4(1)(1)$ $\newline$ Discriminant = $9 - 4$ $\newline$ Discriminant = $5$ $\newline$ The discriminant is positive, so we have two real and distinct solutions for $x$ .
Discriminant Calculation: Apply the quadratic formula to find the two solutions for $x$ . $x = \frac{-(-3) \pm \sqrt{5}}{(2 \cdot 1)}$ $x = \frac{3 \pm \sqrt{5}}{2}$ So, the two solutions for $x$ are $\frac{3 + \sqrt{5}}{2}$ and $\frac{3 - \sqrt{5}}{2}$ .
Solving Quadratic Equation: Now we need to find the value of $x^{3}(x^{3}-18)$ for each solution of $x$ . Let's first consider $x = \frac{3 + \sqrt{5}}{2}$ . We will calculate $x^3$ and then use it to find $x^{3}(x^{3}-18)$ . $x^3 = \left(\frac{3 + \sqrt{5}}{2}\right)^3$ This involves expanding the cube of a binomial.
First Solution Calculation: Expand the cube of the binomial $(3 + \sqrt{5})/2$ .
$x^3 = [(3 + \sqrt{5})/2]^3$
$x^3 = (3/2 + \sqrt{5}/2)^3$
$x^3 = (3/2)^3 + 3*(3/2)^2*(\sqrt{5}/2) + 3*(3/2)*(\sqrt{5}/2)^2 + (\sqrt{5}/2)^3$
This step involves binomial expansion and simplification.
Second Solution Calculation: Simplify the expression for $x^3$ .
$x^3 = \frac{27}{8} + 3\left(\frac{9}{4}\right)\left(\frac{\sqrt{5}}{2}\right) + 3\left(\frac{3}{2}\right)\left(\frac{5}{4}\right) + \frac{5\sqrt{5}}{8}$
$x^3 = \frac{27}{8} + \frac{27\sqrt{5}}{16} + \frac{45}{16} + \frac{5\sqrt{5}}{8}$
$x^3 = \left(\frac{27 + 45}{16}\right) + \left(\frac{27\sqrt{5} + 10\sqrt{5}}{16}\right)$
$x^3 = \frac{72}{16} + \frac{37\sqrt{5}}{16}$
$x^3 = \frac{9}{2} + \frac{37\sqrt{5}}{16}$
Now we have the value of $x^3$ for the first solution.
Correcting Approach: Repeat the process for the second solution $x = \frac{3 - \sqrt{5}}{2}$ . $x^3 = \left(\frac{3 - \sqrt{5}}{2}\right)^3$ This involves expanding the cube of a binomial, similar to the previous steps.
Correcting Approach: Repeat the process for the second solution $x = \frac{3 - \sqrt{5}}{2}$ .
$x^3 = \left[\frac{3 - \sqrt{5}}{2}\right]^3$
This involves expanding the cube of a binomial, similar to the previous steps.Expand the cube of the binomial $\frac{3 - \sqrt{5}}{2}$ .
$x^3 = \left[\frac{3 - \sqrt{5}}{2}\right]^3$
$x^3 = \left(\frac{3}{2} - \frac{\sqrt{5}}{2}\right)^3$
$x^3 = \left(\frac{3}{2}\right)^3 - 3\left(\frac{3}{2}\right)^2\left(\frac{\sqrt{5}}{2}\right) + 3\left(\frac{3}{2}\right)\left(\frac{\sqrt{5}}{2}\right)^2 - \left(\frac{\sqrt{5}}{2}\right)^3$
This step involves binomial expansion and simplification.
Correcting Approach: Repeat the process for the second solution $x = \frac{3 - \sqrt{5}}{2}$ .
$x^3 = \left[\frac{3 - \sqrt{5}}{2}\right]^3$
This involves expanding the cube of a binomial, similar to the previous steps.Expand the cube of the binomial $\frac{3 - \sqrt{5}}{2}$ .
$x^3 = \left[\frac{3 - \sqrt{5}}{2}\right]^3$
$x^3 = \left(\frac{3}{2} - \frac{\sqrt{5}}{2}\right)^3$
$x^3 = \left(\frac{3}{2}\right)^3 - 3\left(\frac{3}{2}\right)^2\left(\frac{\sqrt{5}}{2}\right) + 3\left(\frac{3}{2}\right)\left(\frac{\sqrt{5}}{2}\right)^2 - \left(\frac{\sqrt{5}}{2}\right)^3$
This step involves binomial expansion and simplification.Simplify the expression for $x^3$ .
$x^3 = \frac{27}{8} - 3\left(\frac{9}{4}\right)\left(\frac{\sqrt{5}}{2}\right) + 3\left(\frac{3}{2}\right)\left(\frac{5}{4}\right) - 5\frac{\sqrt{5}}{8}$
$x^3 = \frac{27}{8} - \frac{27\sqrt{5}}{16} + \frac{45}{16} - \frac{5\sqrt{5}}{8}$
$x^3 = \frac{27 + 45}{16} - \frac{27\sqrt{5} + 10\sqrt{5}}{16}$
$x^3 = \left[\frac{3 - \sqrt{5}}{2}\right]^3$ $0$
$x^3 = \left[\frac{3 - \sqrt{5}}{2}\right]^3$ $1$
Now we have the value of $x^3$ for the second solution.
Correcting Approach: Repeat the process for the second solution $x = \frac{3 - \sqrt{5}}{2}$ .
$x^3 = \left(\frac{3 - \sqrt{5}}{2}\right)^3$
This involves expanding the cube of a binomial, similar to the previous steps.Expand the cube of the binomial $\frac{3 - \sqrt{5}}{2}$ .
$x^3 = \left(\frac{3 - \sqrt{5}}{2}\right)^3$
$x^3 = \left(\frac{3}{2} - \frac{\sqrt{5}}{2}\right)^3$
$x^3 = \left(\frac{3}{2}\right)^3 - 3\left(\frac{3}{2}\right)^2\left(\frac{\sqrt{5}}{2}\right) + 3\left(\frac{3}{2}\right)\left(\frac{\sqrt{5}}{2}\right)^2 - \left(\frac{\sqrt{5}}{2}\right)^3$
This step involves binomial expansion and simplification.Simplify the expression for $x^3$ .
$x^3 = \frac{27}{8} - 3\left(\frac{9}{4}\right)\left(\frac{\sqrt{5}}{2}\right) + 3\left(\frac{3}{2}\right)\left(\frac{5}{4}\right) - 5\frac{\sqrt{5}}{8}$
$x^3 = \frac{27}{8} - \frac{27\sqrt{5}}{16} + \frac{45}{16} - \frac{5\sqrt{5}}{8}$
$x^3 = \frac{27 + 45}{16} - \frac{27\sqrt{5} + 10\sqrt{5}}{16}$
$x^3 = \left(\frac{3 - \sqrt{5}}{2}\right)^3$ $0$
$x^3 = \left(\frac{3 - \sqrt{5}}{2}\right)^3$ $1$
Now we have the value of $x^3$ for the second solution.Now we need to find the value of $x^3 = \left(\frac{3 - \sqrt{5}}{2}\right)^3$ $3$ for each solution of $x^3 = \left(\frac{3 - \sqrt{5}}{2}\right)^3$ $4$ .
However, we realize that we made a mistake in the approach. We do not need to find the individual solutions for $x^3 = \left(\frac{3 - \sqrt{5}}{2}\right)^3$ $4$ to calculate $x^3 = \left(\frac{3 - \sqrt{5}}{2}\right)^3$ $3$ because we can use the given equation $x^3 = \left(\frac{3 - \sqrt{5}}{2}\right)^3$ $7$ to simplify the expression directly.
We need to correct our approach and use the given equation to simplify $x^3 = \left(\frac{3 - \sqrt{5}}{2}\right)^3$ $3$ without finding the individual values of $x^3 = \left(\frac{3 - \sqrt{5}}{2}\right)^3$ $4$ .

If x(x−3)=−1 x(x-3)=-1 x(x−3)=−1 then the value of x3(x3−18) x^{3}\left(x^{3}-18\right) x3(x3−18) will be,

If $x(x-3)=-1$ then the value of $x^{3}\left(x^{3}-18\right)$ will be,