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Sagot :
Answer:
2nd option is the correct answer
[tex] 12{b}^{2} \bigg(\sqrt[3]{2a} \bigg)[/tex]
Step-by-step explanation:
[tex]3 {b}^{2} \bigg( \sqrt[3]{54a} \bigg) + 3 \bigg( \sqrt[3]{2a {b}^{6} } \bigg) \\ \\ = 3 {b}^{2} \bigg( \sqrt[3]{27 \times 2a} \bigg) + 3 \bigg( \sqrt[3]{2a {( {b}^{2} )}^{3} } \bigg) \\ \\ = 3 {b}^{2} \bigg( \sqrt[3]{ {3}^{3} \times 2a} \bigg) + 3 \bigg({\sqrt[3]{{( {b}^{2} )}^{3}2a} } \bigg) \\ \\ = 3 {b}^{2} \times 3 \sqrt[ 3]{2a} + 3 {b}^{2} \sqrt[3]{2a} \\ \\ = 9 {b}^{2} \sqrt[ 3]{2a} + 3 {b}^{2} \sqrt[3]{2a} \\ \\ = 12{b}^{2} \bigg(\sqrt[3]{2a} \bigg)[/tex]
Answer:
second option: 12b²(∛2a)
Step-by-step explanation:
you can break [tex]\sqrt[3]{54a}[/tex] into [tex]\sqrt[3]{3^3(2a)}[/tex] which equals 3[tex]\sqrt[3]{2a}[/tex]
you can break [tex]\sqrt[3]{2ab^6}[/tex] into [tex]\sqrt[3]{2ab^3b^3}[/tex] which equals b²[tex]\sqrt[3]{2a}[/tex]
now we can simplify: (3b² · 3[tex]\sqrt[3]{2a}[/tex]) + 3b²[tex]\sqrt[3]{2a}[/tex] = 9b²[tex]\sqrt[3]{2a}[/tex] + 3b²[tex]\sqrt[3]{2a}[/tex]
this equals 12b²[tex]\sqrt[3]{2a}[/tex]
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