Pre calculus fraction simplify questionSimplify: 16x481−y42x3+yWolfram alpha confirms the answer from the answer sheet: Wolframalpha answer

Let u=2x3 and notice that u4=(2x3)4=24x434=16x481So we get:16x481−y42x3+y=u4−y4u+y=(u2−y2)(u2+y2)u+y=(u−y)(u+y)(u2+y2)u+y=(u−y)(u2+y2)=(2x3−y)(4x29+y2)=(13(2x−3y))(19(4x2+9y2))=127(2x−3y)(4x2+9y2)=127(8x3−12x2y+18xy2−27y3)

Multiplying by 81 on the top and the bottom gives16x4−81y427(2x+3y)=(4x2−9y2)(4x2+9y2)27(2x+3y)=(2x−3y)(2x+3y)(4x2+9y2)27(2x+3y)=127(2x−3y)(4x2+9y2)

Let u=2x3 and notice that u4=(2x3)4=24x434=16x481So we get:16x481−y42x3+y=u4−y4u+y=(u2−y2)(u2+y2)u+y=(u−y)(u+y)(u2+y2)u+y=(u−y)(u2+y2)=(2x3−y)(4x29+y2)=(13(2x−3y))(19(4x2+9y2))=127(2x−3y)(4x2+9y2)=127(8x3−12x2y+18xy2−27y3)

Multiplying by 81 on the top and the bottom gives16x4−81y427(2x+3y)=(4x2−9y2)(4x2+9y2)27(2x+3y)=(2x−3y)(2x+3y)(4x2+9y2)27(2x+3y)=127(2x−3y)(4x2+9y2)

Solved: "Pre calculus fraction simplify questionSimplify: 16x481−y42x3+yWolfram alpha confirms..."

Alejandro Atkins

Answered question

2022-04-30

Pre calculus fraction simplify question
Simplify:

\frac{\frac{16 x^{4}}{81} - y^{4}}{\frac{2 x}{3} + y}

Wolfram alpha confirms the answer from the answer sheet: Wolframalpha answer

Answer & Explanation

bailaretzy33

Beginner2022-05-01Added 15 answers

Let

u = \frac{2 x}{3}

and notice that

u^{4} = {(\frac{2 x}{3})}^{4} = \frac{2^{4} x^{4}}{3^{4}} = \frac{16 x^{4}}{81}

So we get:

\frac{\frac{16 x^{4}}{81} - y^{4}}{\frac{2 x}{3} + y} = \frac{u^{4} - y^{4}}{u + y} = \frac{(u^{2} - y^{2}) (u^{2} + y^{2})}{u + y} = \frac{(u - y) (u + y) (u^{2} + y^{2})}{u + y} = (u - y) (u^{2} + y^{2}) = (\frac{2 x}{3} - y) (\frac{4 x^{2}}{9} + y^{2})

= (\frac{1}{3} (2 x - 3 y)) (\frac{1}{9} (4 x^{2} + 9 y^{2})) = \frac{1}{27} (2 x - 3 y) (4 x^{2} + 9 y^{2}) = \frac{1}{27} (8 x^{3} - 12 x^{2} y + 18 x y^{2} - 27 y^{3})

haillarip0c9

Beginner2022-05-02Added 23 answers

Multiplying by 81 on the top and the bottom gives

\frac{16 x^{4} - 81 y^{4}}{27 (2 x + 3 y)} = \frac{(4 x^{2} - 9 y^{2}) (4 x^{2} + 9 y^{2})}{27 (2 x + 3 y)} = \frac{(2 x - 3 y) (2 x + 3 y) (4 x^{2} + 9 y^{2})}{27 (2 x + 3 y)} =

\frac{1}{27} (2 x - 3 y) (4 x^{2} + 9 y^{2})

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