The given expression is ![3b^2*(\sqrt[3]{54a}) + 3*(\sqrt[3]{2ab^6})](https://tex.z-dn.net/?f=%203b%5E2%2A%28%5Csqrt%5B3%5D%7B54a%7D%29%20%2B%203%2A%28%5Csqrt%5B3%5D%7B2ab%5E6%7D%29%20)
This can be simplified as :
= ![3*b^2*(\sqrt[3]{27 *2*a}) + 3*(\sqrt[3]{2*a*b^6})](https://tex.z-dn.net/?f=%203%2Ab%5E2%2A%28%5Csqrt%5B3%5D%7B27%20%2A2%2Aa%7D%29%20%2B%203%2A%28%5Csqrt%5B3%5D%7B2%2Aa%2Ab%5E6%7D%29%20)
We know that: ![\sqrt[3]{27} = 3](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B27%7D%20%20%3D%203%20%20%20)
Similarly we also can simplify: ![\sqrt[3]{b^6} = b^2](https://tex.z-dn.net/?f=%20%20%5Csqrt%5B3%5D%7Bb%5E6%7D%20%20%3D%20b%5E2%20)
So our expression will look like this:
= ![3*3*b^2*(\sqrt[3]{2a}) + 3*b^2*(\sqrt[3]{2a})](https://tex.z-dn.net/?f=%203%2A3%2Ab%5E2%2A%28%5Csqrt%5B3%5D%7B2a%7D%29%20%2B%203%2Ab%5E2%2A%28%5Csqrt%5B3%5D%7B2a%7D%29%20)
= ![9b^2*(\sqrt[3]{2a}) + 3b^2*(\sqrt[3]{2a})](https://tex.z-dn.net/?f=%209b%5E2%2A%28%5Csqrt%5B3%5D%7B2a%7D%29%20%2B%203b%5E2%2A%28%5Csqrt%5B3%5D%7B2a%7D%29%20)
=![\sqrt[3]{2a}*(9b^2 + 3b^2)](https://tex.z-dn.net/?f=%20%20%5Csqrt%5B3%5D%7B2a%7D%2A%289b%5E2%20%2B%203b%5E2%29%20)
=![\sqrt[3]{2a}*(12b^2)](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B2a%7D%2A%2812b%5E2%29%20)
This can also be written as:
![12b^2(\sqrt[3]{2a})](https://tex.z-dn.net/?f=%2012b%5E2%28%5Csqrt%5B3%5D%7B2a%7D%29%20)
So the Answer is Option B
3/4 + 1/12 = 10/12 and then you simplify it and get 5/6 as your final answer.
Answer:
what? this isn't a true/false question...
the answer is 6
Answer:
X = 0, π/2 in the interval [0, 2pi).
Step-by-step explanation:
Use the auxiliary angle method:
R sin(x + a) = Rsin x cos a + Rcos x sin a = 1
sin x + cos x = 1
Comparing coefficients:
R cos a = 1 and R sin a = 1, so
tan a = R sin a / R cos a = 1
So a = π/4 radians.
Also R^2(sin^2 a + cos^2 a) = 1^2 + 1^2 = 2
Therefore R = √2.
So √2 sin (x +π/4 = 1
sin x + π/4 = 1/√2
x + π/4 = π/4
x = 0 radians
Also
x = 0 + π/2 = π/2.
