What is the cube root of -700×2×49×5

Answer 1

you can pull out a 7, as the 7 and 49 multiply to 7^3 leaving 100 x 2 x 5

and pull out a 10 as 10^3 = 100 x 2 x 5

cube root is 7x 10 = 70

but you're left with 70 times cube root of -1

cube root of -1= (1/2)(1 + or - i(sqr3)

70(1/2)(1+ or - isqr3) = 35 + or - isqr3 = 35 + or - 60.6i

sqr3 = approximately 1.732

a+bi = cube root of -1

cube both sides

(a+bi)^3 = -1

expand

a^3 +3a^2bi + 3a(bi)^2 + (bi)^3 = -1

a^3 +3a^2bi -3ab^2 -b^3 = -1

set the two real terms = -1

a^3-3ab^2 =-1

set the two imaginary terms = 0

3a^2b-b^3 =0

factor out b

b(3a^2 - b^2) =0

3a^2 = b^2

substitute 3a^2 for b^2 in the real terms equation

a^3 -3ab^2 =-1

a^3-3a(3a^2) =-1

a^3 -9a^3= -1

-8a^3 =-1

a^3 = 1/8

a = 1/2

b^2 = 3a^2 = 3(1/2)^2 = 3/4

b= + or -(1/2)sqr3

cube root of -1 = (1/2)(1 + or - isqr3)

70 x (1/2)(1+ or - isqr3)

= 35 + or - 35isqr3

=35 + or - 60.6i