Certainly! Let's walk through the evaluation of the expression [tex]\( 3 \sqrt[3]{16} + 3 \sqrt[3]{54} - 3 \sqrt[3]{128} \)[/tex] step-by-step.
1. Calculate the cube root of 16:
[tex]\[
\sqrt[3]{16}
\][/tex]
The cube root of 16 is approximately [tex]\( 2.5198421 \)[/tex].
2. Multiply the cube root of 16 by 3:
[tex]\[
3 \times \sqrt[3]{16} = 3 \times 2.5198421 = 7.5595263
\][/tex]
3. Calculate the cube root of 54:
[tex]\[
\sqrt[3]{54}
\][/tex]
The cube root of 54 is approximately [tex]\( 3.7797631 \)[/tex].
4. Multiply the cube root of 54 by 3:
[tex]\[
3 \times \sqrt[3]{54} = 3 \times 3.7797631 = 11.3392894
\][/tex]
5. Calculate the cube root of 128:
[tex]\[
\sqrt[3]{128}
\][/tex]
The cube root of 128 is approximately [tex]\( 5.0396842 \)[/tex].
6. Multiply the cube root of 128 by 3:
[tex]\[
3 \times \sqrt[3]{128} = 3 \times 5.0396842 = 15.1190526
\][/tex]
7. Sum the first two products and subtract the third:
[tex]\[
7.5595263 + 11.3392894 - 15.1190526
\][/tex]
8. Calculate the final result:
[tex]\[
7.5595263 + 11.3392894 = 18.8988157
\][/tex]
[tex]\[
18.8988157 - 15.1190526 = 3.7797631
\][/tex]
Therefore, the result of the expression [tex]\( 3 \sqrt[3]{16} + 3 \sqrt[3]{54} - 3 \sqrt[3]{128} \)[/tex] is approximately [tex]\( 3.7797631 \)[/tex].
Let's summarize the intermediate values and the final result as follows:
1. The term [tex]\( 3 \sqrt[3]{16} \)[/tex] evaluates to approximately [tex]\( 7.5595263 \)[/tex].
2. The term [tex]\( 3 \sqrt[3]{54} \)[/tex] evaluates to approximately [tex]\( 11.3392894 \)[/tex].
3. The term [tex]\( 3 \sqrt[3]{128} \)[/tex] evaluates to approximately [tex]\( 15.1190526 \)[/tex].
4. The final result of the expression is approximately [tex]\( 3.7797631 \)[/tex].
