Question

factorisation
$9a ^{4 } - 24a {}^{2}b {}^{2} + 16b {}^{4} - 25b$
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Answer 1

9a⁴ — 24a²b² + 16b⁴ — 256

= (9a⁴ — 24a²b² + 16b⁴) — 256

= [(3a²)² — 2 x 3a² x 4b² + (4b²)²] —162

= (3a² —402)² — 162

= [(3a² – 4b²) -16] [(3a² – 42) + 16]

= (3a² – 4b² -16) (3a² – 4b² + 16)

Hope it will help you

Answer 2

$9a^4 - 24a^2b^2 + 16b^4 - 25b$

=$(3a^2)^2- 2×3a^24b^2 + (4b^2)^2 - 25b$

=$(3a^2 - 4b^2)^2 - (5rootb)^2$

=$(3a^2 - 4b^2 + 5rootb)(3a^2 - 4b^2 - 5rootb)$

so answer is

=$(3a^2 - 4b^2 + 5rootb)(3a^2 - 4b^2 - 5rootb)$

if we consider 25b as 256
then 256 = 16^2
so 5rootb will be replaced by 16