An example

An example is playfully for grade 6 from Math and I don't know how to explain it to my daughter when I don't want to use the calculator to calculate the cube root. Thus:

A cuboid was made from a block of 16x18x48 mm of modeline. What will be the edge of the cube? If I calculate the cuboid volume, would I have to cube root or not? Thank you for your answer.

Correct result:

a2 =  24 mm

Solution:

a=16 mm b=18 mm c=48 mm  V=a b c=16 18 48=13824 mm3 16=24 18=2 32 48=24 3  13824=29×33=(23)3 33=83 33=(8 3)3=243  V=a23 x0=a+b+c3=16+18+48382327.3333 mm V0=273=19683 mm3 V1=283=21952 mm3 V3=263=17576 mm3 V4=253=15625 mm3 V5=243=13824 mm3 V5=V a2=24 mm   Correctness test:  13824=29 33=(23 3)3=(8 3)3=243 a2=24  a2=V3=138243=24 mma=16 \ \text{mm} \ \\ b=18 \ \text{mm} \ \\ c=48 \ \text{mm} \ \\ \ \\ V=a \cdot \ b \cdot \ c=16 \cdot \ 18 \cdot \ 48=13824 \ \text{mm}^3 \ \\ 16=2^4 \ \\ 18=2 \cdot \ 3^2 \ \\ 48=2^4 \cdot \ 3 \ \\ \ \\ 13824=2^9 \times 3^3=(2^3)^3 \cdot \ 3^3=8^3 \cdot \ 3^3=(8 \cdot \ 3)^3=24^3 \ \\ \ \\ V=a_{2}^3 \ \\ x_{0}=\dfrac{ a+b+c }{ 3 }=\dfrac{ 16+18+48 }{ 3 } \doteq \dfrac{ 82 }{ 3 } \doteq 27.3333 \ \text{mm} \ \\ V_{0}=27^3=19683 \ \text{mm}^3 \ \\ V_{1}=28^3=21952 \ \text{mm}^3 \ \\ V_{3}=26^3=17576 \ \text{mm}^3 \ \\ V_{4}=25^3=15625 \ \text{mm}^3 \ \\ V_{5}=24^3=13824 \ \text{mm}^3 \ \\ V_{5}=V \ \\ a_{2}=24 \ \text{mm} \ \\ \ \\ \text{ Correctness test: } \ \\ 13824=2^9 \cdot \ 3^3=( 2^3 \cdot \ 3)^3=(8 \cdot \ 3)^3=24^3 \ \\ a_{2}=24 \ \\ \ \\ a_{2}=\sqrt[3]{ V}=\sqrt[3]{ 13824 }=24 \ \text{mm}



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