Question

if a^4+16=4a^2 then what is the value os a^6+64

Answer 1

a⁴ + 16 = 4a²
(a²)² + (4)² = 4a²
(a² + 4)² - 8a² = 4a²
(a² + 4)² = 12a²
(a² + 4) = ±2√3a

now , a^6 + 64 = ?
a^6 + 64
= (a²)³ + (4)³
=(a² + 4)² - 3.4.a²(a² +4)
=(a² + 4)² - 12a²(a² + 4)

case 1 :-
put (a² + 4) = 2√3a

=(2√3a)³ -12a2(2√3a)
=24√3a³ - 24√3a³ = 0

again,
case 2 :-
put (a² + 4) = -2√3a

=( -2√3a)³ -12a²(-2√3a)
= -24√3a³ + 24√3a³
= 0

we take a^6 + 64 = 0 in both cases
hence, (a^6 + 64) = 0

Answer 2

a⁴ + 16 = 4a² 
(a²)² + (4)² = 4a² 
(a² + 4)² - 8a² = 4a² 
(a² + 4)² = 12a² 
(a² + 4) = ±2√3a 
a⁶ + 64 
= (a²)³ + (4)³ 
=(a² + 4)² - 3.4.a²(a² +4) 
=(a² + 4)² - 12a²(a² + 4) 
Again, 1st time
by put (a² + 4) = 2√3a 
=(2√3a)³ -12a2(2√3a)
=24√3a³ - 24√3a³ = 0

again, 2nd time
put (a² + 4) = -2√3a 
=( -2√3a)³ -12a²(-2√3a)
= -24√3a³ + 24√3a³ 
= 0 
taking a⁶ + 64 = 0 in both 1st time ans 2nd time
∴ (a⁶ + 64) = 0