If a^{3}+b^{3}+c^{3}=8072,a:b=b:c=3:2 then the value of a is

AP ICET 2017 Shift 1 Paper

Question : 78 of 200

Marks: +1, -0

a^{3}+b^{3}+c^{3}=8072,a:b=b:c=3:2

then the value of a is

Solution:

Let

a=9k,b=6k,c=4k

⇒(9k)^3+(6k)^3+(4k)^3=8072

⇒k^3(729+216+64)=8072

⇒k^3=8072/1009=8

⇒k=2

∴a=9k=18

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