Question

Find a cube of 2.1.

Answer 1

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.

Complete step by step answer:

  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