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## Find the cube roots 8(cos 216° + i sin 216°)?

[8(cos 216° + i sin 216°)] ^ ⅓

I use a sneaky application of DeMoivre’s theorem to find roots, even though it’s normally used to find powers. The sneaky part is to just add 360°k or 2πk to the angle. Therefore, the cube roots are:

8^(1/3) [cos ⅓(216°+360k) + i sin ⅓(216°+360k) ]

= 2[cos (72° + 120°k) + i sin (72° + 120°k)]

Start at k = 0, 1, 2, because there are THREE cube roots of a complex number:

k = 0: 2(cos 72° + i sin 72°)

k = 1: 2(cos 192° + i sin 192°)

k = 2: 2(cos 312° + i sin 312°)

Find the cube-roots of : -216 x 1728 – Sarthaks eConnect – cubes and cube roots. class-8. 0 votes. cubes and cube roots. icse. class-8. 0 votes. 1 answer. Find the cube-roots of : -216. asked Sep 16, 2018 in Mathematics by AnujPatel (53.4k points).Use this calculator to find the cube root of positive or negative numbers. The cube root of x is the same as x raised to the 1/3 power. Cube root of 216 is 6.Free math problem solver answers your precalculus homework questions with step-by-step explanations.

Cube Root Calculator – where . So the third roots are. thanks!!!! find the length of side AB give your answer to 1 decimal place.Mathematics NCERT Grade 8, Chapter 7: Cubes and Cube Roots- The chapter starts by citing an interesting story about India's great mathematician S.Ramanujan. Solved examples show how students need to find the cube root of any number step-wise. The chapter ends with a summary.Proving Trigonometric Identity – (Sin squared + cos squared = 1). Finding areas with trigonometry. The meaning of i. Patterns with imaginary numbers. Rational or irrational. Root 2 is Irrational – Proof by contradiction. Finding the cube roots of 8.

Mathway | Precalculus Problem Solver – To take the cube root of each of these, take the cube root of the 125 ( which is 5) and divide the argument of the exponential by 3 (raising to the 1/3 power). This results in 5 ei 96 ; 5 ei 216 ; 5 ei 336 Now using DeMovrie (Euler formula) one gets.The cube roots `z_c` of `z` are then all complex numbers with argument `(4pi)/15` and modulus 2 Write each expression in the standard form for a complex number, a + bi. a. [3(cos(27°)) + isin(27°) z1=2(cos(pi/5)+isin(pi/5)) and z2=8(cos(7pi/6)+sin(7pi/6))How do I calculate for z1z2, z2 , z1 using…How do I find the roots of $\sqrt[3]{ – 11 – 2i}$ ? Tried to use Moivre's theorem, but can not find the solutions by using the polar form Remark: These are also the solutions which you obtain when resolving the quadratic in $x$ and then compute the third root of these solutions.