resolve into factors.
please solve this question
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1)The factors of a⁶ + b⁶ is (a² + b²) (a⁴ - a²b² + b⁴).
2)x
6
−y
6
=(x
2
)
3
−(y
2
)
3
=(x
2
−y
2
)[(x
2
)
2
+(y
2
)
2
+(x
2
×y
2
)]=(x−y)(x+y)(x
2
+y
2
+xy)(x
2
+y
2
−xy)
(using identities a
2
−b
2
=(a+b)(a−b) and a
4
+b
4
+a
2
b
2
=(a
2
+b
2
+ab)(a
2
+b
2
−ab) )
Hence, x
6
−y
6
=(x−y)(x+y)(x
2
+y
2
+xy)(x
2
+y
2
−xy)
3)First of all see there is x³ and 8y²³ where 8y²³ can be written as 2³•y²³ or, as(2y)²³.and now if you compare it to a³-b³=(a-b)(a²-ab+b²),keeping in mind that a=x,b=2y.Then you get x³-8y²³=(x-2)(x²-2xy+4y²)
4)2y(y²+3ײ)
5)1+9a+26a²+28a³
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