Question

If a is a real number such that a3 + 4a - 8 = 0, then find the value of a7 + 64a2.

Answer 1

Answer:

$a^7+64a^2=128$

Step-by-step explanation:

Formula used:

$(a-b)^2=a^2+b^2-2ab$

Given:

$a^3+4a-8=0$

$a^3+4a=8$...............(1)

$a^3=8-4a$

squaring on both sides

$(a^3)^2=(8-4a)^2$

$a^6=64+16a^2-64a$

Multiply both sides by 'a'

we get

$a^7=64a+16a^3-64a^2$

$a^7+64a^2=64a+16a^3$

$a^7+64a^2=64a+16a^3$

$a^7+64a^2=16(4a+a^3)$

$a^7+64a^2=16(a^3+4a)$

$a^7+64a^2=16(8)$ (by using (1) )

$a^7+64a^2=128$

Answer 2

Answer:

Refer to the above answer