Factorise: x^3+3x^2y+3xy^2-125
Answers
Answer:
x
3
+3x
2
y+3xy
2
+y
3
−125=0
⇒(x
3
+3x
2
y+3xy
2
+y
3
)=125
⇒[x
3
+3xy(x+y)+y
3
]=125
Using the identity, a
3
+3ab(a+b)+b
3
=(a+b)
3
we have
x
3
+3xy(x+y)+y
3
=(x+y)
3
⇒(x+y)
3
=5
3
⇒(x+y)
3
−5
3
is of the form a
3
−b
3
=(a−b)(a
2
+ab+b
2
)
=(x+y−5)((x+y)
2
+5(x+y)+5
2
)
=(x+y−5)(x
2
+y
2
+2xy+5x+5y+25)
Step-by-step explanation:
Hope it will help u
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Factorise:
x
3
+3x
2
y+3xy
2
+y
3
−125=0
Medium
Solution
verified
Verified by Toppr
x
3
+3x
2
y+3xy
2
+y
3
−125=0
⇒(x
3
+3x
2
y+3xy
2
+y
3
)=125
⇒[x
3
+3xy(x+y)+y
3
]=125
Using the identity, a
3
+3ab(a+b)+b
3
=(a+b)
3
we have
x
3
+3xy(x+y)+y
3
=(x+y)
3
⇒(x+y)
3
=5
3
⇒(x+y)
3
−5
3
is of the form a
3
−b
3
=(a−b)(a
2
+ab+b
2
)
=(x+y−5)((x+y)
2
+5(x+y)+5
2
)
=(x+y−5)(x
2
+y
2
+2xy+5x+5y+25)