Question
Answer
$$108t^6+(9t^2+a)^2a-(3t^2+a)^2(12t^2+a)=0$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}108t^6+(9t^2+a)^2a-(3t^2+a)^2(12t^2+a)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}108t^6+(81t^4+18at^2+a^2)a-(9t^4+6at^2+a^2)(12t^2+a) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}108t^6+81at^4+18a^2t^2+a^3-(108t^6+9at^4+72at^4+6a^2t^2+12a^2t^2+a^3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}108t^6+81at^4+18a^2t^2+a^3-(108t^6+81at^4+18a^2t^2+a^3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}108t^6+81at^4+18a^2t^2+a^3-108t^6-81at^4-18a^2t^2-a^3 \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{108t^6}+ \cancel{81at^4}+ \cancel{18a^2t^2}+ \cancel{a^3} -\cancel{108t^6} -\cancel{81at^4} -\cancel{18a^2t^2} -\cancel{a^3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}0\end{aligned} $$ | |
| ① | Find $ \left(9t^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}{ 9t^2 } $ and $ B = \color{red}{ a }$. $$ \begin{aligned}\left(9t^2+a\right)^2 = \color{blue}{\left( 9t^2 \right)^2} +2 \cdot 9t^2 \cdot a + \color{red}{a^2} = 81t^4+18at^2+a^2\end{aligned} $$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 }$. $$ \begin{aligned}\left(3t^2+a\right)^2 = \color{blue}{\left( 3t^2 \right)^2} +2 \cdot 3t^2 \cdot a + \color{red}{a^2} = 9t^4+6at^2+a^2\end{aligned} $$ |
| ② | $$ \left( \color{blue}{81t^4+18at^2+a^2}\right) \cdot a = 81at^4+18a^2t^2+a^3 $$ Multiply each term of $ \left( \color{blue}{9t^4+6at^2+a^2}\right) $ by each term in $ \left( 12t^2+a\right) $. $$ \left( \color{blue}{9t^4+6at^2+a^2}\right) \cdot \left( 12t^2+a\right) = 108t^6+9at^4+72at^4+6a^2t^2+12a^2t^2+a^3 $$ |
| ③ | Combine like terms: $$ 108t^6+81at^4+18a^2t^2+a^3 = 108t^6+81at^4+18a^2t^2+a^3 $$ |
| ④ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 108t^6+81at^4+18a^2t^2+a^3 \right) = -108t^6-81at^4-18a^2t^2-a^3 $$ |
| ⑤ | Combine like terms: $$ \, \color{blue}{ \cancel{108t^6}} \,+ \, \color{green}{ \cancel{81at^4}} \,+ \, \color{blue}{ \cancel{18a^2t^2}} \,+ \, \color{green}{ \cancel{a^3}} \, \, \color{blue}{ -\cancel{108t^6}} \, \, \color{green}{ -\cancel{81at^4}} \, \, \color{blue}{ -\cancel{18a^2t^2}} \, \, \color{green}{ -\cancel{a^3}} \, = 0 $$ |
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