Question 51

Find the smallest integer whose cube is equal to itself.

Solution

Solve from the options one by one,

I.$$ {1}$$--> $$1^3$$ = $$ {1}$$

II. $${2}$$--> $$2^3$$ = $$ {8}$$

III. $${-1}$$-->$$-1^3$$ = $$ {-1}$$

IV. $${0}$$-->$$0^3$$ = $$ {0}$$

In the case $$ {1}$$,$$ {-1}$$ and $$ {0}$$, the cube of the number is itself but among these three numbers $$ {-1}$$ is the smallest number.

therefore $$ {-1}$$ is the answer

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