Solve a^3-{[(1/2a-4/3r)(9/2a-12r)-(3/4a+2r)(-3a+8r)](frac{2}{9}a-frac{8}{3}r)+4ar(frac{11}{3}a-8r)}

$$a ^ { 3 } - \{ [ ( \frac { 1 } { 2 } a - \frac { 4 } { 3 } r ) ( \frac { 9 } { 2 } a - 12 r ) - ( \frac { 3 } { 4 } a + 2 r ) ( - 3 a + 8 r ) ] ( \frac { 2 } { 9 } a - \frac { 8 } { 3 } r ) + 4 a r ( \frac { 11 } { 3 } a - 8 r ) \}$$

Evaluate

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Solution Steps

Use the distributive property to multiply \frac{1}{2}a-\frac{4}{3}r by \frac{9}{2}a-12r and combine like terms.

$$a^{3}-\left(\left(\frac{9}{4}a^{2}-12ar+16r^{2}-\left(\frac{3}{4}a+2r\right)\left(-3a+8r\right)\right)\left(\frac{2}{9}a-\frac{8}{3}r\right)+4ar\left(\frac{11}{3}a-8r\right)\right)$$

Use the distributive property to multiply \frac{3}{4}a+2r by -3a+8r and combine like terms.

$$a^{3}-\left(\left(\frac{9}{4}a^{2}-12ar+16r^{2}-\left(-\frac{9}{4}a^{2}+16r^{2}\right)\right)\left(\frac{2}{9}a-\frac{8}{3}r\right)+4ar\left(\frac{11}{3}a-8r\right)\right)$$

To find the opposite of -\frac{9}{4}a^{2}+16r^{2}, find the opposite of each term.

$$a^{3}-\left(\left(\frac{9}{4}a^{2}-12ar+16r^{2}+\frac{9}{4}a^{2}-16r^{2}\right)\left(\frac{2}{9}a-\frac{8}{3}r\right)+4ar\left(\frac{11}{3}a-8r\right)\right)$$

Combine \frac{9}{4}a^{2} and \frac{9}{4}a^{2} to get \frac{9}{2}a^{2}.

$$a^{3}-\left(\left(\frac{9}{2}a^{2}-12ar+16r^{2}-16r^{2}\right)\left(\frac{2}{9}a-\frac{8}{3}r\right)+4ar\left(\frac{11}{3}a-8r\right)\right)$$

Combine 16r^{2} and -16r^{2} to get 0.

$$a^{3}-\left(\left(\frac{9}{2}a^{2}-12ar\right)\left(\frac{2}{9}a-\frac{8}{3}r\right)+4ar\left(\frac{11}{3}a-8r\right)\right)$$

Use the distributive property to multiply \frac{9}{2}a^{2}-12ar by \frac{2}{9}a-\frac{8}{3}r and combine like terms.

$$a^{3}-\left(a^{3}-\frac{44}{3}a^{2}r+32ar^{2}+4ar\left(\frac{11}{3}a-8r\right)\right)$$

Use the distributive property to multiply 4ar by \frac{11}{3}a-8r.

$$a^{3}-\left(a^{3}-\frac{44}{3}a^{2}r+32ar^{2}+\frac{44}{3}a^{2}r-32ar^{2}\right)$$

Combine -\frac{44}{3}a^{2}r and \frac{44}{3}a^{2}r to get 0.

$$a^{3}-\left(a^{3}+32ar^{2}-32ar^{2}\right)$$

Combine 32ar^{2} and -32ar^{2} to get 0.

$$a^{3}-a^{3}$$

Combine a^{3} and -a^{3} to get 0.

$$0$$

Factor

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Solution Steps

Factor out \frac{1}{18}.

$$\frac{18a^{3}-\left(\left(\left(3a-8r\right)\left(3a-8r\right)-\left(3a+8r\right)\left(-3a+8r\right)\right)\left(a-12r\right)+24ar\left(11a-24r\right)\right)}{18}$$

Simplify.

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