Posted by AMY on Friday, January 14, 2011 at 8:43am.
what happened to the f^6 ?
(If you expanded your answer would you get the original ?)
16s^6-2f^6
= 2(8s^6 - f^6)
now 8s^6 = (2s^2)^3 and f^6 = (f^2)^3 so
knowing that A^3 - B^3 = (A-B)(a^2 + AB + b^2)
2(8s^6 - f^6)
= 2(2s^2 - f^2)(4s^4 + 2s^2f^2 + f^4)
How can 2s - 16 be the same as 16s^6-2f^6 ? Those functions are totally different. One is sixth order in s and is linear in s. One has f as a variable and the other does not.
16s^6 -2f^6 = 2(8s^6 - f^6)
= 2*(sqrt8 s^3 -f^3)(sqrt8 s^3+ f^3)
Further factoring of the cubic terms is possible. Look up how to factor a^3 - b^3 and a^3 + b^3.
thank u
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