Question
Answer
$$4t^6-(3t^2-a)^2a-(t^2-a)^2(4t^2-a)=0$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}4t^6-(3t^2-a)^2a-(t^2-a)^2(4t^2-a)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4t^6-(9t^4-6at^2+a^2)a-(1t^4-2at^2+a^2)(4t^2-a) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4t^6-(9at^4-6a^2t^2+a^3)-(4t^6-at^4-8at^4+2a^2t^2+4a^2t^2-a^3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}4t^6-(9at^4-6a^2t^2+a^3)-(4t^6-9at^4+6a^2t^2-a^3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}4t^6-9at^4+6a^2t^2-a^3-(4t^6-9at^4+6a^2t^2-a^3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}4t^6-9at^4+6a^2t^2-a^3-4t^6+9at^4-6a^2t^2+a^3 \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{4t^6} -\cancel{9at^4}+ \cancel{6a^2t^2} -\cancel{a^3} -\cancel{4t^6}+ \cancel{9at^4} -\cancel{6a^2t^2}+ \cancel{a^3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}0\end{aligned} $$ | |
| ① | Find $ \left(3t^2-a\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 3t^2 } $ and $ B = \color{red}{ a }$. $$ \begin{aligned}\left(3t^2-a\right)^2 = \color{blue}{\left( 3t^2 \right)^2} -2 \cdot 3t^2 \cdot a + \color{red}{a^2} = 9t^4-6at^2+a^2\end{aligned} $$Find $ \left(t^2-a\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ t^2 } $ and $ B = \color{red}{ a }$. $$ \begin{aligned}\left(t^2-a\right)^2 = \color{blue}{\left( t^2 \right)^2} -2 \cdot t^2 \cdot a + \color{red}{a^2} = t^4-2at^2+a^2\end{aligned} $$ |
| ② | $$ \left( \color{blue}{9t^4-6at^2+a^2}\right) \cdot a = 9at^4-6a^2t^2+a^3 $$ Multiply each term of $ \left( \color{blue}{t^4-2at^2+a^2}\right) $ by each term in $ \left( 4t^2-a\right) $. $$ \left( \color{blue}{t^4-2at^2+a^2}\right) \cdot \left( 4t^2-a\right) = 4t^6-at^4-8at^4+2a^2t^2+4a^2t^2-a^3 $$ |
| ③ | Combine like terms: $$ 4t^6 \color{blue}{-at^4} \color{blue}{-8at^4} + \color{red}{2a^2t^2} + \color{red}{4a^2t^2} -a^3 = 4t^6 \color{blue}{-9at^4} + \color{red}{6a^2t^2} -a^3 $$ |
| ④ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 9at^4-6a^2t^2+a^3 \right) = -9at^4+6a^2t^2-a^3 $$ |
| ⑤ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 4t^6-9at^4+6a^2t^2-a^3 \right) = -4t^6+9at^4-6a^2t^2+a^3 $$ |
| ⑥ | Combine like terms: $$ \, \color{blue}{ \cancel{4t^6}} \, \, \color{green}{ -\cancel{9at^4}} \,+ \, \color{blue}{ \cancel{6a^2t^2}} \, \, \color{green}{ -\cancel{a^3}} \, \, \color{blue}{ -\cancel{4t^6}} \,+ \, \color{green}{ \cancel{9at^4}} \, \, \color{blue}{ -\cancel{6a^2t^2}} \,+ \, \color{green}{ \cancel{a^3}} \, = 0 $$ |
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