If $ a+2 b=5; $ then show that : $ a^{3}+8 b^{3}+30 a b=125 $.
Given,
\( a+2 b=5 ; \)
We have to show that
\( a^{3}+8 b^{3}+30 a b=125 \)
\( a+2 b=5 \)
Cubing both sides:
\( (a+2 b)^{3}=(5)^{3} \)
\( \Rightarrow a^{3}+(2 b)^{3}+\{3 a \times 2 b(a+2 b)\}=(5)^{3} \)
\( \Rightarrow a^{3}+8 b^{3}+6 a b(a+2 b)=125 \)
\( \Rightarrow a^{3}+8 b^{3}+6 a b(5)=125 \)
\( ( \) as, \( \quad a+2 b=5) \)
\( \Rightarrow a^{3}+8 b^{3}+30 a b=125 \)
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