TS ICET 23 May 2019 Shift 2 Paper

Question : 79 of 200

Marks: +1, -0

If for the natural numbers a, b, c,

a^3+ b^3 + c^3

= 8072 and a: b = b: c = 3:2, then a =

Solution:

a : b =3: 2, b : c =3: 2

= a : b =9: b\;\;d : c =6: 4

a : b : c =9: 6: 4

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

a^{3}+b^{3}+c^{3}=8072

(9 k)^{3}+(6 k)^{3}+(4 k)^{3}=8072

1009 k^{3}=8072

k^{3}=8 ⇒k=2

∴a=9 k=9 × 2=18

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