Question

# Find a cube of 2.1.

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Hint: A cube root of a no. x is no. y such that ${y^3}{\text{ }}-x$ all non –zero real no have exactly one real cube roots are a pair of complex conjugate cube roots and all non-zero complex no have three distinct complex cube roots.
The cube root of a no. x is the no. y which satisfy the e.g. ${y^3} = x$
The cube root formula is used to give the cube root value of any No. is found by multiplying the number three times.
e.g. $5 \times 5 \times 5 = 125$
Now, the cube root formula is the vice versa of the cube formula.

As discussed in the hint part cube of 2.1 will be two times multiplying the No. with itself.
Step 1: 1st multiplying the no. itself
i.e. $2.1 \times 2.1 = 4.41$
step 2: now multiplying the step1st no. with 2.1
i.e. $4.41 \times 2.1 = 9.261$
Cube of $2.1$ is$9.261$.

Note: The cube root of a no. is a special value that when cubed gives the original No. the cube gives the original no. The cube root of 27 is 3 because when 3 is cubed you get 27. the cubed of a no. n is its third power that is the result of multiplying three instances of n together with the cube of no. or any other mathematical expression is devoted by a subscript 3
For e.g. ${2^3} = 8$ or $\left( {x{\text{ }} + 1} \right)$
The cube is also the No. multiplied by its square.
${N^3} =$$n \times n$2 $=$ $n \times n \times n$
the cube function is the function x-x3 that maps a No. to its cube .it is an odd function as
${\left( { - n} \right)^3} = {\text{ }} - {\text{ }}({n^3}).$
every $+ ve$ rational No. is the sum of three$+ ve$ rational cubes and there are rational that are not sum of the two rational cubes occasionally have the subjective property in other field such as $q{\text{ }} = \;fp$ for such that prime p that $p \ne 1$ (mod$3$) . but not necessarily see the counterexample with rationales above