12+ X^3-1 Background. Thank you for the help: (1 + a2)3 = 13 + 3.12.a2 + 3.1.(a2)2 + (a2)3 = 1 + 3a2 + 3a4 + a6.
Think of a number you can put into that equation for x and get it to equal 0.
Misnotation forgetting those brackets involved in 3 we resolve thereat that x=2 when b=2.5/3, but when we divide by x, to yeild the x+(1/x)=3b we get x. If you want the real solution of your initial equation, then you simply take the real cubic root of $u$, and then $v$ is uniquely defined by $v=\dfrac 1{3u}$ (or you can also take the real cubic root of $v$ for. (1 + a2)3 = 13 + 3.12.a2 + 3.1.(a2)2 + (a2)3 = 1 + 3a2 + 3a4 + a6. Precalculus complex numbers in trigonometric form powers of complex numbers.