Math

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Factor:
9a²b⁴-4c²+16b²c²-72a²b⁷.
Solution
Look at the expression:
9a²b⁴-4c²+16b²c²-72a²b⁷.
Group the first term with the last, the second — with the third.
9a²b⁴-4c²+16b²c²-72a²b⁷=(9a²b⁴-72a²b⁷)+(-4c²+16b²c²).
Factor brackets by carrying out common divisor, minus:
(9a²b⁴-72a²b⁷)+(-4c²+16b²c²)=9a²b⁴(1-8b³)-4c²(1-4b²).
Remark that 1-8b³=1³-2³b³=1³-(2b)³, so we can factor the expression (1-8b³) using the formula of two cubes' difference:
p³-q³=(p-q)(p²+pq+q²).
Remark that 1-4b²=1²-2²b²=1²-(2b)², so we can factor the expression (1-4b²) using the formula of two squares' difference:
m²-n²=(m-n)(m+n).
We get:
9a²b⁴(1-8b³)-4c²(1-4b²)=9a²b⁴(1-2b)(1+2b+4b²)-4c²(1-2b)(1+2b).
Factor the last expression by carrying out the common divisor (1-2b):
9a²b⁴(1-2b)(1+2b+4b²)-4c²(1-2b)(1+2b)= (1-2b)(9a²b⁴(1+2b+4b²)-4c²(1+2b)).
Expand brackets standing in larger bracket:
(1-2b)(9a²b⁴(1+2b+4b²)-4c²(1+2b))=(1-2b)(9a²b⁴+18a²b⁵+36a²b⁶-4c²-8bc²).
9a²b⁴-4c²+16b²c²-72a²b⁷=(1-2b)(9a²b⁴+18a²b⁵+36a²b⁶-4c²-8bc²).
The puzzle is solved.