What is (sqrt(1^2+(2t)^2+((3t^2)/4)^2))^3 ?

The solution to (sqrt(1^2+(2t)^2+((3t^2)/4)^2))^3 is ((16+64t^2+9t^4)sqrt(16+64t^2+9t^4))/(64)

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simplify (sqrt(1^2+(2t)^2+((3t^2)/4)^2))^3

simplify

1^{2} + (2 t)^{2} + (\frac{3 t ^{2}}{4})^{2}^{3}

\frac{( 16 + 64 t ^{2} + 9 t ^{4} ) 16 + 64 t ^{2} + 9 t ^{4}}{64}

Solution steps

1^{2} + (2 t)^{2} + (\frac{3 t ^{2}}{4})^{2}^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

1^{2} + (2 t)^{2} + (\frac{3 t ^{2}}{4})^{2}^{2 + 1} = 1^{2} + (2 t)^{2} + (\frac{3 t ^{2}}{4})^{2}^{2} 1^{2} + (2 t)^{2} + (\frac{3 t ^{2}}{4})^{2}

= 1^{2} + (2 t)^{2} + (\frac{3 t ^{2}}{4})^{2}^{2} 1^{2} + (2 t)^{2} + (\frac{3 t ^{2}}{4})^{2}

Apply exponent rule:

assuming

n i s o dd or a \geq 0

1^{2} + (2 t)^{2} + (\frac{3 t ^{2}}{4})^{2}^{2} = 1^{2} + (2 t)^{2} + (\frac{3 t ^{2}}{4})^{2}

= (1^{2} + (2 t)^{2} + (\frac{3 t ^{2}}{4})^{2}) 1^{2} + (2 t)^{2} + (\frac{3 t ^{2}}{4})^{2}

1^{2} = 1

(2 t)^{2} = 4 t^{2}

(\frac{3 t ^{2}}{4})^{2} = \frac{9 t ^{4}}{16}

= (1 + 4 t^{2} + \frac{9 t ^{4}}{16}) 1^{2} + (2 t)^{2} + (\frac{3 t ^{2}}{4})^{2}

Simplify

1^{2} + (2 t)^{2} + (\frac{3 t ^{2}}{4})^{2} : \frac{16 + 64 t ^{2} + 9 t ^{4}}{4}

= (1 + 4 t^{2} + \frac{9 t ^{4}}{16}) \frac{16 + 64 t ^{2} + 9 t ^{4}}{4}

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 1 + 4 t ^{2} + \frac{9 t ^{4}}{16} ) 16 + 64 t ^{2} + 9 t ^{4}}{4}

Simplify

(1 + 4 t^{2} + \frac{9 t ^{4}}{16}) 16 + 64 t^{2} + 9 t^{4} : \frac{( 16 + 64 t ^{2} + 9 t ^{4} ) 16 + 64 t ^{2} + 9 t ^{4}}{16}

= \frac{\frac{( 16 + 64 t ^{2} + 9 t ^{4} ) 16 + 64 t ^{2} + 9 t ^{4}}{16}}{4}

Apply the fraction rule:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

= \frac{( 16 + 64 t ^{2} + 9 t ^{4} ) 16 + 64 t ^{2} + 9 t ^{4}}{16 \cdot 4}

Multiply the numbers:

16 \cdot 4 = 64

= \frac{( 16 + 64 t ^{2} + 9 t ^{4} ) 16 + 64 t ^{2} + 9 t ^{4}}{64}

(y + 2)^{3} (y - 2)^{3}

2 p \cdot (- 3) q \cdot (- 4) p q^{2}

\frac{1}{s - 3} e^{- (s - 3) t}

6 (2^{5})

What is (sqrt(1^2+(2t)^2+((3t^2)/4)^2))^3 ?
The solution to (sqrt(1^2+(2t)^2+((3t^2)/4)^2))^3 is ((16+64t^2+9t^4)sqrt(16+64t^2+9t^4))/(64)