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Byju's Answer
Standard IX
Mathematics
Algebraic Identities
Factorise.1 x...
Question
Factorise.
(1) x
3
+ 64y
3
(2) 125p
3
+ q
3
(3) 125k
3
+ 27m
3
(4) 2l
3
+ 432m
3
(5) 24a
3
+ 81b
3
(6) y
3
+
1
8
y
3
(7) a
3
+
8
a
3
(8) 1 +
q
3
125
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Solution
It is known that,
a
3
+
b
3
=
a
+
b
a
2
+
b
2
-
a
b
1
x
3
+
64
y
3
=
x
3
+
4
y
3
=
x
+
4
y
x
2
+
4
y
2
-
x
×
4
y
=
x
+
4
y
x
2
+
16
y
2
-
4
x
y
2
125
p
3
+
q
3
=
5
p
3
+
q
3
=
5
p
+
q
5
p
2
+
q
2
-
5
p
×
q
=
5
p
+
q
25
p
2
+
q
2
-
5
p
q
3
125
k
3
+
27
m
3
=
5
k
3
+
3
m
3
=
5
k
+
3
m
5
k
2
+
3
m
2
-
5
k
×
3
m
=
5
k
+
3
m
25
k
2
+
9
m
2
-
15
k
m
4
2
l
3
+
432
m
3
=
2
l
3
+
216
m
3
=
2
l
3
+
6
m
3
=
2
l
+
6
m
l
2
+
6
m
2
-
l
×
6
m
=
2
l
+
6
m
l
2
+
36
m
2
-
6
l
m
5
24
a
3
+
81
b
3
=
3
8
a
3
+
27
b
3
=
3
2
a
3
+
3
b
3
=
3
2
a
+
3
b
2
a
2
+
3
b
2
-
2
a
×
3
b
=
3
2
a
+
3
b
4
a
2
+
9
b
2
-
6
a
b
6
y
3
+
1
8
y
3
=
y
3
+
1
2
y
3
=
y
+
1
2
y
y
2
+
1
2
y
2
-
y
×
1
2
y
=
y
+
1
2
y
y
2
+
1
4
y
2
-
1
2
7
a
3
+
8
a
3
=
a
3
+
2
a
3
=
a
+
2
a
a
2
+
2
a
2
-
a
×
2
a
=
a
+
2
a
a
2
+
4
a
2
-
2
8
1
+
q
3
125
=
1
3
+
q
5
3
=
1
+
q
5
1
2
+
q
5
2
-
1
×
q
5
=
1
+
q
5
1
+
q
2
25
-
q
5
Suggest Corrections
4
Similar questions
Q.
Factorise. (i)
125
p
3
+
q
3
(ii)
24
a
3
+
81
b
3
Q.
Factories. (1 )
x
3
+
64
y
3
(2)
24
a
3
+
81
b
3
Q.
If
√
1
−
x
6
+
√
1
−
y
6
=
a
3
.
(
x
3
−
y
3
)
, prove that
d
y
d
x
=
x
2
y
2
√
1
−
y
6
1
−
x
6
Q.
Factorise :
24
a
3
+
37
a
2
−
5
a
Q.
Factorise:
x
3
+
3
x
2
y
+
3
x
y
2
+
y
3
−
125
=
0
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