Answer:
[tex]2a^3b^2\sqrt[3]{3a}[/tex]
Step-by-step explanation:
Use the following rules for exponents:
[tex]a^m*a^n=a^{m+n}\\\\\sqrt[3]{x^3}=x[/tex]
Simplify 24. Find two factors of 24, one of which should be a perfect cube:
[tex]8*3=24\\\\2^3=8[/tex]
Insert:
[tex]\sqrt[3]{2^3*3a^{10}b^6}[/tex]
Now split the exponents. Split 10 into as many 3's as possible:
[tex]10=3+3+3+1[/tex]
Insert as exponents:
[tex]\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^6}[/tex]
Split 6 into as many 3's as possible:
[tex]6=3+3[/tex]
Insert as exponents:
[tex]\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^3*b^3}[/tex]
Now simplify. Any terms with an exponent of 3 will be moved out of the radical (rule #2):
[tex]2\sqrt[3]{3*a^3*a^3*a^3*a^1*b^3*b^3}\\\\\\2*a*a*a\sqrt[3]{3*a^1*b^3*b^3}\\\\\\2*a*a*a*b*b\sqrt[3]{3*a^1}[/tex]
Simplify:
[tex]2a^3b^2\sqrt[3]{3a}[/tex]
:Done
