$\bf \textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
(a+b)(a^2-ab+b^2)= a^3+b^3 \\ \quad \\
a^3-b^3 = (a-b)(a^2+ab+b^2)\qquad
(a-b)(a^2+ab+b^2)= a^3-b^3\\\\
-------------------------------\\\\
\boxed{64=4^3}\qquad x^3+64\implies x^3+4^3\implies (x+4)(x^2-x4+4^2)
\\\\\\
(x+4)(x^2-4x+16)$

8 0

3 years ago

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7x+5y=-13<br> -2x=7y+26<br> Solve by elimination

RSB [31]

Answer:

Point form : (1,-4)

Equation for, : x=1, y=-4

Step-by-step explanation: Add the fractions in order to solve for the first variable. Plug this vale into the other equations in order to solve for the remaining variables.

Hope this helps you out! ☺

5 0

3 years ago

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Solve y In terms of x

miss Akunina [59]

the answer is $y=3x-6$

subtract 6x on both sides → $-2y=-6x+12$

divide both sides by -2y → $y=3x-6$

4 0

3 years ago

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