8b²+c²=(2b+c²)(4b²-2bc²+c4)
Answers
The factorise the given expression a^2-4b^2+a^3-8b^3-(a-2b)^2a
2
−4b
2
+a
3
−8b
3
−(a−2b)
2
is (a-2b)(4b+a^{2}+2ab+4b^{2}).(a−2b)(4b+a
2
+2ab+4b
2
).
Step-by-step explanation:
We have,
a^2-4b^2+a^3-8b^3-(a-2b)^2a
2
−4b
2
+a
3
−8b
3
−(a−2b)
2
To factorise the given expression=?
=(a^2-4b^2)+(a^3-8b^3)-(a-2b)^2=(a
2
−4b
2
)+(a
3
−8b
3
)−(a−2b)
2
=[a^2-(2b)^2]+[a^3-(2b)^3]-(a-2b)^2=[a
2
−(2b)
2
]+[a
3
−(2b)
3
]−(a−2b)
2
=(a+2b)(a-2b)+(a-2b)(a^{2}+2ab+4b^{2})-(a-2b)^2=(a+2b)(a−2b)+(a−2b)(a
2
+2ab+4b
2
)−(a−2b)
2
Using identities,
a^{2}-b^{2} =(a+b)(a-b)a
2
−b
2
=(a+b)(a−b) and
a^{3}-b^{3} =(a-b)(a^{2}+ab+b^{2})a
3
−b
3
=(a−b)(a
2
+ab+b
2
)
Taking (a-2b) as common, we get
=(a-2b)[(a+2b)+(a^{2}+2ab+4b^{2})-(a-2b)]=(a−2b)[(a+2b)+(a
2
+2ab+4b
2
)−(a−2b)]
=(a-2b)(a+2b+a^{2}+2ab+4b^{2}-a+2b)=(a−2b)(a+2b+a
2
+2ab+4b
2
−a+2b)
=(a-2b)(4b+a^{2}+2ab+4b^{2})=(a−2b)(4b+a
2
+2ab+4b
2
)
=(a-2b)(4b+a^{2}+2ab+4b^{2})=(a−2b)(4b+a
2
+2ab+4b
2
)
Hence, the factorise the given expression a^2-4b^2+a^3-8b^3-(a-2b)^2a
2
−4b
2
+a
3
−8b
3
−(a−2b)
2
is (a-2b)(4b+a^{2}+2ab+4b^{2}).(a−2b)(4b+a
2
+2ab+4b
2