Solve the given inequality algebraically.

$x^3>1$

The given inequality is : $x^3>1$

We have to solve for x.

Zeroes of the inequality $f\left(x\right)=x^3-1>0$are

$x^3-1=0$
localid="1646169167170" role="math" $(x-1)(x^2+x+1)=0$

So we get $x=1$

$(x^2+x+1)$has no real solutions

Therefore$x=1$

Now we use the zeroes to separate the real number line into intervals$(-\infty ,1)(1,\infty )$

Now we select a test number in each interval found in Step 3 and evaluate at each number to determine if $f\left(x\right)=x^3-1=0$is positive or negative

In the interval $(-\infty ,1)$we chose 0 where f is negative

In the interval $(1,\infty )$we chose 2 where f is positive

We know that our required inequality is $f\left(x\right)>0$

Here the inequality is not strict $(\ge or\le )$so we have to exclude the solutions of $f\left(x\right)=0$in the solution set.

So we want the interval where f is positive.

So required solution set is$(1,\infty )$