Solution 2.1:1g

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The expression in the exercise is of the form <math> (a-b)^2 </math>, where <math> a=y^2</math> and <math> b=3x^2 </math>. With the help of the squaring rule <math> (a-b)^2 =a^2 -2ab +b^2 </math>, we have
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{{Displayed math||<math>\begin{align}
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(y^2-3x^3)^2 &= (y^2)^2 -2\cdot y^2\cdot 3x^3 +(3x^3)^2 \\[3pt]
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&= y^{2\cdot 2} -6x^3y^2 +3^2x^{3\cdot 2}\\[3pt]
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&= y^4 -6x^3y^2 +9x^6\\[3pt]
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&= 9x^6 -6x^3y^2 +y^4\,\textrm{.}
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\end{align}</math>}}

Current revision

The expression in the exercise is of the form \displaystyle (a-b)^2 , where \displaystyle a=y^2 and \displaystyle b=3x^2 . With the help of the squaring rule \displaystyle (a-b)^2 =a^2 -2ab +b^2 , we have

\displaystyle \begin{align}

(y^2-3x^3)^2 &= (y^2)^2 -2\cdot y^2\cdot 3x^3 +(3x^3)^2 \\[3pt] &= y^{2\cdot 2} -6x^3y^2 +3^2x^{3\cdot 2}\\[3pt] &= y^4 -6x^3y^2 +9x^6\\[3pt] &= 9x^6 -6x^3y^2 +y^4\,\textrm{.} \end{align}

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