Question #68e59

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Question #68e59

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Answer 1 · 2017-09-12T13:43:31

$\frac{2 a^{3} + b^{3}}{a^{3} + 2 b^{3}}$ $(2a^3 + b^3) / (a^3+2b^3)$

Explanation:

$\frac{{(a + b)}^{3} + a^{3}}{{(a + b)}^{3} + b^{3}}$ $((a+b)^3 + a^3) / ((a+b)^3 + b^3)$

A rule of exponents is that variables with an operation within the bracket are raised to the exponent. E.g.
${(p + q)}^{2}$ $(p+q)^2$
$= p^{2} + q^{2}$ $=p^2 + q^2$

So here, you apply the same rule:
$= \frac{a^{3} + b^{3} + a^{3}}{a^{3} + b^{3} + b^{3}}$ $=(a^3 + b^3 + a^3) / (a^3 + b^3 + b^3)$
$= \frac{2 a^{3} + b^{3}}{a^{3} + 2 b^{3}}$ $=(2a^3 + b^3) / (a^3 + 2b^3)$

You can't factorise the expression because of the +

Hope this helps!

Answer 2 · 2017-09-12T14:38:30

$\frac{2 a + b}{2 b + a}$ $(2a+b)/(2b+a)$

Explanation:

${(a + b)}^{3} + a^{3} is a sum of cubes$ $(a+b)^3+a^3" is a "color(blue)"sum of cubes"$

$∙ x x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2}) \leftarrow factors$ $•color(white)(x)x^3+y^3=(x+y)(x^2-xy+y^2)larrcolor(blue)" factors"$

$here x = a + b and y = a$ $"here "x=a+b" and "y=a$

$\Rightarrow {(a + b)}^{3} + a^{3}$ $rArr(a+b)^3+a^3$

$= (a + b + a) ({(a + b)}^{2} - a (a + b) + a^{2})$ $=(a+b+a)((a+b)^2-a(a+b)+a^2)$

$=(2a+b)(a^2+2ab+b^2cancel(-a^2)-abcancel(+a^2))$

$=(2a+b)(a^2+b^2+ab)$

$color(blue)"Similarly for denominator"$

$"here "x=a+b" and " y=b$

$rArr(a+b)^3+b^3$

$=(a+b+b)((a+b)^2-b(a+b)+b^2)$

$=(2b+a)(a^2+2ab+b^2-abcancel(-b^2)cancel(+b^2))$

$=(2b+a)(a^2+b^2+ab)$

$"Putting it together we obtain"$

$((2a+b)cancel((a^2+b^2+ab)))/((2b+a)cancel((a^2+b^2+ab)))$

$=(2a+b)/(2b+a)$

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Question #68e59

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