Question
Answer
$$sqrt\dfrac{(x^2-1)(x-1)}{9y^2z^{16}}=\dfrac{qrstx^3-qrstx^2-qrstx+qrst}{9y^2z^{16}}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}sqrt\frac{(x^2-1)(x-1)}{9y^2z^{16}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}sqrt\frac{x^3-x^2-x+1}{9y^2z^{16}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{qrstx^3-qrstx^2-qrstx+qrst}{9y^2z^{16}}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x^2-1}\right) $ by each term in $ \left( x-1\right) $. $$ \left( \color{blue}{x^2-1}\right) \cdot \left( x-1\right) = x^3-x^2-x+1 $$ |
| ② | Multiply $qrst$ by $ \dfrac{x^3-x^2-x+1}{9y^2z^{16}} $ to get $ \dfrac{ qrstx^3-qrstx^2-qrstx+qrst }{ 9y^2z^{16} } $. Step 1: Write $ qrst $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} qrst \cdot \frac{x^3-x^2-x+1}{9y^2z^{16}} & \xlongequal{\text{Step 1}} \frac{qrst}{\color{red}{1}} \cdot \frac{x^3-x^2-x+1}{9y^2z^{16}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ qrst \cdot \left( x^3-x^2-x+1 \right) }{ 1 \cdot 9y^2z^{16} } \xlongequal{\text{Step 3}} \frac{ qrstx^3-qrstx^2-qrstx+qrst }{ 9y^2z^{16} } \end{aligned} $$ |
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