Question
Answer
$$\dfrac{(a^2+b^2)^3-3a^2(a^2+b^2)^2}{{(a^2+b^2)^3}^2}=\dfrac{a^6+3a^4b^2+3a^2b^4+b^6-(3a^6+6a^4b^2+3a^2b^4)}{{(a^2+b^2)^3}^2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(a^2+b^2)^3-3a^2(a^2+b^2)^2}{{(a^2+b^2)^3}^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{(a^2+b^2)^3-3a^2(1a^4+2a^2b^2+b^4)}{{(a^2+b^2)^3}^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{a^6+3a^4b^2+3a^2b^4+b^6-(3a^6+6a^4b^2+3a^2b^4)}{{(a^2+b^2)^3}^2}\end{aligned} $$ | |
| ① | Find $ \left(a^2+b^2\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ a^2 } $ and $ B = \color{red}{ b^2 }$. $$ \begin{aligned}\left(a^2+b^2\right)^2 = \color{blue}{\left( a^2 \right)^2} +2 \cdot a^2 \cdot b^2 + \color{red}{\left( b^2 \right)^2} = a^4+2a^2b^2+b^4\end{aligned} $$ |
| ② | Find $ \left(a^2+b^2\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = a^2 $ and $ B = b^2 $. $$ \left(a^2+b^2\right)^3 = \left( a^2 \right)^3+3 \cdot \left( a^2 \right)^2 \cdot b^2 + 3 \cdot a^2 \cdot \left( b^2 \right)^2+\left( b^2 \right)^3 = a^6+3a^4b^2+3a^2b^4+b^6 $$ |
| ③ | Multiply $ \color{blue}{3a^2} $ by $ \left( a^4+2a^2b^2+b^4\right) $ $$ \color{blue}{3a^2} \cdot \left( a^4+2a^2b^2+b^4\right) = 3a^6+6a^4b^2+3a^2b^4 $$ |
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