$\bf \textit{difference and sum of cubes}
\\\\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
(a+b)(a^2-ab+b^2)= a^3+b^3 
\\\\
a^3-b^3 = (a-b)(a^2+ab+b^2)\qquad
(a-b)(a^2+ab+b^2)= a^3-b^3\\\\
-------------------------------\\\\
64g^3+8\implies 4^3g^3+8\implies (4g)^3+2^3
\\\\\\
(4g+2)[(4g)^2-(4g)(2)+2^2]\implies (4g+2)[(4^2g^2)-8g+4]
\\\\\\
(4g+2)(16g^2-8g+4)$

3 0

3 years ago

Combine the like terms to create an equivalent expression:<br> −3x-6+(-1)

asambeis [7]

Answer:

-3x - 7

Step-by-step explanation:

-3x - 6 - 1 = -3x - 7

3 0

2 years ago

Read 2 more answers

Niña jumps 225 times in 3 minutes. At this rate how many times can she jump in 5 minutes

Tresset [83]

225/3 = 75
75 x 5 = 375

Hope this helps.

5 0

3 years ago

Read 2 more answers

How do you solve #10

IgorLugansk [536]

$LINEAR \: \: EQUATIONS \\ \\ \\ Given \: expression \: \: - \: \\ \\ \frac{7}{3} (2x + 3) + \frac{3}{4} ( \frac{x}{5} - \frac{15}{2} ) = \frac{11}{8} \\ \\ \frac{14x}{3} + \frac{21}{3} + \frac{3x}{20} - \frac{45}{8} = \frac{11}{8} \\ \\ \frac{14x}{3} + \frac{3x}{20} + \frac{21}{3} - \frac{45}{8} = \frac{11}{8} \\ \\ \frac{289x}{60} = \frac{11}{8} + \frac{45}{8} - \frac{21}{3} \\ \\ \frac{289x}{60} = \frac{56}{8} - \frac{21}{3} \\ \\ \frac{289x}{60} = 7 - 7 = 0 \\ \\ \frac{289x}{60} = 0 \\ \\ \\ \: \: \: \:|| \: \: \: x = 0 \: \: \:|| \: \: \: \: \: \: \: \: \: \: \: \: \: Ans.$

3 0

2 years ago