X^3+3x^2y+3xy^2+y^3 ———————————— X^3+y^3

1 answer:

7 0

To factor this fraction, you have be be aware of two special factoring formula:
a^3 + b^3 = (a + b)(a^2 – ab + b^2)

(a+b)³ = a³ + 3a²b + 3ab² + b³

You can see the top part in this case is (x+y)^3, and the bottom (denominator) can be factor into (x+y)(x^2-xy+y^2)
we can cancel (x+y), so what we have left is (x+y)^2/(x^2-xy+y^2)
or (x^2+2xy+y^2)/(x^2-xy+y^2)

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Answer:

Proved Below

Step-by-step explanation:

$\sf \frac{3}{m+4} - \frac{4}{m} = 6\\LCM = m(m+4)\\Multiplying \ both \ sides \ by\ m(m+4)\\3m - 4(m+4) = 6m(m+4)\\3m-4m-16 = 6m^2+24m\\-m-16 = 6m^2+24m\\Adding\ m\ to\ both\ sides\\-16 = 6m^2+24m+m\\-16 = 6m^2+25m\\Adding \ 16 \ to \ both \ sides\\0 = 6m^2+25m+16\\OR\\6m^2+25m+16 = 0$

Hence, Proved that $\sf \frac{3}{m+4} - \frac{4}{m} = 6$ is equivalent to $\sf 6m^2+25m+16 = 0$ .

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