Is there a number that is exactly 3 more than its cube [Solved]

x = x^3 3
x^3 x 3 = 0

Let f(x)=x^3 x 3
f(-2)=-3
f(-1)=3
Hence there is a value of x between x=-2 and x=-1 which satisfies the specified condition. In fact, this is the one value (do a plot of the graph).

f(x) = x^3 x 3.

So consider the function

I dont think there is a whole number

So consider the function

Now f(-2) = -3 < 0 and f(0) = 3 > 0 so by the intermediate value theorem there exists at least one real number a, with -2 < a < 0 such that f(a) = 0. That number 'a', is the required number.

x = x^3 3
x^3 x 3 = 0
x ~ -1.6717
Its not a rational number, and the expression for it is quite complicated.

-1/6*(324 12*717^(1/2))^(1/3) -2/(324 12*717^(1/2))^(1/3) or roughly -1.671699882,
1/12*(324 12*717^(1/2))^(1/3) 1/((324 12*717^(1/2))^(1/3)) ( or -)1/2*I*3^(1/2)*(-1/6*(324 12*717^(1/2))^(1/3) 2/(324 12*717^(1/2))^(1/3)) or roughly 0.8358499409( or -)1.046869319*I (where I=sqrt(-1))

x = x^3 3
x^3 x 3 = 0
x ~ -1.6717
Its not a rational number, and the expression for it is quite complicated.

This post is last updated on hrtanswers.com at Date : 1st of September – 2022

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