NMTC

Algebra Level 3

If a a is a real number such that a 3 + 4 a 8 = 0 a^3+4a-8=0 , then find the value of a 7 + 64 a 2 a^7+64a^2 .


The answer is 128.

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2 solutions

Aaryan Maheshwari
Oct 11, 2017

a 3 + 4 a 8 = 0 a^3+4a-8=0 a 3 = 4 ( a 2 ) \Rightarrow\space a^3=-4(a-2) a 6 = ( a 3 ) 2 = 16 ( a 2 + 4 4 a ) = 16 a 2 + 64 64 a \Rightarrow\space a^6=(a^3)^2=16(a^2+4-4a)=16a^2+64-64a a 7 = a ( a 6 ) = a ( 16 a 2 + 64 64 a ) = 16 a 3 + 64 a 64 a 2 \Rightarrow\space a^7=a(a^6)=a(16a^2+64-64a)=16a^3+64a-64a^2 a 7 + 64 a 2 = 16 a 3 + 64 a = 16 ( a 3 + 4 a ) = 16 × 8 = 128. \therefore\space a^7+64a^2=16a^3+64a=16(a^3+4a)=16\times8=128.

Chew-Seong Cheong
Oct 11, 2017

a 3 + 4 a 8 = 0 a 7 + 4 a 5 8 a 4 = 0 a 7 + 4 a 3 ( a 2 2 a ) = 0 a 7 + 4 ( 4 a + 8 ) ( a 2 2 a ) = 0 a 7 16 a ( a 2 ) 2 = 0 a 7 16 a ( a 2 4 a + 4 ) = 0 a 7 16 a 3 + 64 a 2 64 a = 0 a 7 + 64 a 2 = 16 ( a 3 + 4 a ) = 16 ( 4 a + 8 + 4 a ) = 128 \begin{aligned} a^3 + 4a-8 &=0 \\ a^7 + 4a^5-8a^4 &=0 \\ a^7 + 4a^3(a^2-2a) &=0 \\ a^7 + 4(-4a+8)(a^2-2a) &=0 \\ a^7 - 16a(a-2)^2 &=0 \\ a^7 - 16a(a^2-4a+4) &=0 \\ a^7 - 16a^3 + 64a^2-64a &=0 \\ a^7 + 64a^2 &= 16(a^3 +4a) \\ &= 16(-4a+8+4a) \\ &= \boxed{128} \end{aligned}

you did the same thing, but in a different way. Excellent sir.

Aaryan Maheshwari - 3 years, 8 months ago

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