factorise
16V2 a3 - 31353 – 24V3a²b + 18V2ab2
Answers
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Answer:
bc-2
Step-by-step explanation:
Answer: Factorized form is (\sqrt{2}a+2\sqrt{2}b+c)( 2a^2+8b^2+c^2-4ab-2\sqrt{2} bc-\sqrt{2}ac)(
2
a+2
2
b+c)(2a
2
+8b
2
+c
2
−4ab−2
2
bc−
2
ac)
Step-by-step explanation:
Since we have given that
2\sqrt{2}a^3+16\sqrt{2}b^3+c^3-12abc2
2
a
3
+16
2
b
3
+c
3
−12abc
We need to factorise the above expression:
As we know the identity of cubes:
x^3+y^3+x^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)x
3
+y
3
+x
3
−3xyz=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)
Here,
\begin{gathered}(\sqrt{2}a)^3+(2\sqrt{2}b)^3+c^3-12abc\\\\=(\sqrt{2}a+2\sqrt{2}b+c)((\sqrt{2}a)^2+(2\sqrt{2}b)^2+c^2-\sqrt{2}a\times 2\sqrt{2}b -2\sqrt{2}bc-\sqrt{2}ac)\\\\=(\sqrt{2}a+2\sqrt{2}b+c)( 2a^2+8b^2+c^2-4ab-2\sqrt{2} bc-\sqrt{2}ac)\end{gathered}
(
2
a)
3
+(2
2
b)
3
+c
3
−12abc
=(
2
a+2
2
b+c)((
2
a)
2
+(2
2
b)
2
+c
2
−
2
a×2
2
b−2
2
bc−
2
ac)
=(
2
a+2
2
b+c)(2a
2
+8b
2
+c
2
−4ab−2
2
bc−
2
ac)
Hence, factorized form is (\sqrt{2}a+2\sqrt{2}b+c)( 2a^2+8b^2+c^2-4ab-2\sqrt{2} bc-\sqrt{2}ac)(
2
a+2
2
b+c)(2a
2
+8b
2
+c
2
−4ab−2
2
bc−
2
ac)
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