Докажите тождество:1).(a^2+4)^2-16a^2=(a+2)^2(a-2)^22).(4a+1)^2(4a-1)^2=(16a^2+1)^2-64a^2

Question

Докажите тождество:
1).(a^2+4)^2-16a^2=(a+2)^2(a-2)^2
2).(4a+1)^2(4a-1)^2=(16a^2+1)^2-64a^2

Answer 1

1) (a^2 + 4)^2 - 16a^2 = (a + 2)^2 * (a - 2)^2
a^4 + 8a^2 + 16 - 16a^2 = (a^2 + 4a + 4) * (a^2 - 4a + 4)
a^ - 8a^2 + 16 = a^4 - 4a^3 + 4a^2 + 4a^3 - 16a^2 + 16a + 4a^2-16a - 16
a^4 - 8a^2 + 16 = a^ - 8a^2 + 16
a^ - 8a^2 + 16 -a^4 + 8a^2 - 16 = 0

(4a + 1)^2 * (4a - 1)^2 = (16a^2 + 1)^2 - 64a^2
(16a^2 + 8a +1) * (16a^2 - 8a + 1) = 256a^4 + 32a^2 1 - 64a^2
256a^4 - 128a^3 + 16a^2 + 128a^3 - 64a^2 + 8a + 16a^2 -8a + 1 = 256a^4 - 32a^2 + 1
256a^4 - 32a^2 + 1 = 256a^4 - 32a^2 + 1
256a^4 - 32a^2 + 1 -256a^4 + 32a^2 - 1 = 0