Solve -(-4/3)^{-2}x^{8}+0,75x^{2}(1/3x^{2}+4/3)(1/4x^{4}-x^2)]:(-frac{1}{2}x^2)^2+2(x^4-frac{1}{4})

$$- ( - \frac { 4 } { 3 } ) ^ { - 2 } x ^ { 8 } + 0,75 x ^ { 2 } ( \frac { 1 } { 3 } x ^ { 2 } + \frac { 4 } { 3 } ) ( \frac { 1 } { 4 } x ^ { 4 } - x ^ { 2 } ) ] : ( - \frac { 1 } { 2 } x ^ { 2 } ) ^ { 2 } + 2 ( x ^ { 4 } - \frac { 1 } { 4 } )$$

Evaluate

$$-\frac{x^{8}}{2}-x^{4}$$

Solution Steps

Calculate -\frac{4}{3} to the power of -2 and get \frac{9}{16}.

$$-\frac{9}{16}x^{8}+0.75x^{2}\left(\frac{1}{3}x^{2}+\frac{4}{3}\right)\left(\frac{1}{4}x^{4}-x^{2}\right)$$

Use the distributive property to multiply 0.75x^{2} by \frac{1}{3}x^{2}+\frac{4}{3}.

$$-\frac{9}{16}x^{8}+\left(\frac{1}{4}x^{4}+x^{2}\right)\left(\frac{1}{4}x^{4}-x^{2}\right)$$

Consider \left(\frac{1}{4}x^{4}+x^{2}\right)\left(\frac{1}{4}x^{4}-x^{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$-\frac{9}{16}x^{8}+\left(\frac{1}{4}x^{4}\right)^{2}-\left(x^{2}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$-\frac{9}{16}x^{8}+\left(\frac{1}{4}x^{4}\right)^{2}-x^{4}$$

Expand \left(\frac{1}{4}x^{4}\right)^{2}.

$$-\frac{9}{16}x^{8}+\left(\frac{1}{4}\right)^{2}\left(x^{4}\right)^{2}-x^{4}$$

To raise a power to another power, multiply the exponents. Multiply 4 and 2 to get 8.

$$-\frac{9}{16}x^{8}+\left(\frac{1}{4}\right)^{2}x^{8}-x^{4}$$

Calculate \frac{1}{4} to the power of 2 and get \frac{1}{16}.

$$-\frac{9}{16}x^{8}+\frac{1}{16}x^{8}-x^{4}$$

Combine -\frac{9}{16}x^{8} and \frac{1}{16}x^{8} to get -\frac{1}{2}x^{8}.

$$-\frac{1}{2}x^{8}-x^{4}$$

Expand

$$-\frac{x^{8}}{2}-x^{4}$$

Solution Steps

Calculate -\frac{4}{3} to the power of -2 and get \frac{9}{16}.

$$-\frac{9}{16}x^{8}+0.75x^{2}\left(\frac{1}{3}x^{2}+\frac{4}{3}\right)\left(\frac{1}{4}x^{4}-x^{2}\right)$$

Use the distributive property to multiply 0.75x^{2} by \frac{1}{3}x^{2}+\frac{4}{3}.

$$-\frac{9}{16}x^{8}+\left(\frac{1}{4}x^{4}+x^{2}\right)\left(\frac{1}{4}x^{4}-x^{2}\right)$$

Consider \left(\frac{1}{4}x^{4}+x^{2}\right)\left(\frac{1}{4}x^{4}-x^{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$-\frac{9}{16}x^{8}+\left(\frac{1}{4}x^{4}\right)^{2}-\left(x^{2}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$-\frac{9}{16}x^{8}+\left(\frac{1}{4}x^{4}\right)^{2}-x^{4}$$

Expand \left(\frac{1}{4}x^{4}\right)^{2}.

$$-\frac{9}{16}x^{8}+\left(\frac{1}{4}\right)^{2}\left(x^{4}\right)^{2}-x^{4}$$

To raise a power to another power, multiply the exponents. Multiply 4 and 2 to get 8.

$$-\frac{9}{16}x^{8}+\left(\frac{1}{4}\right)^{2}x^{8}-x^{4}$$

Calculate \frac{1}{4} to the power of 2 and get \frac{1}{16}.

$$-\frac{9}{16}x^{8}+\frac{1}{16}x^{8}-x^{4}$$

Combine -\frac{9}{16}x^{8} and \frac{1}{16}x^{8} to get -\frac{1}{2}x^{8}.

$$-\frac{1}{2}x^{8}-x^{4}$$