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Byju's Answer
Standard X
Mathematics
Factorisation of Quadratic Polynomials
Factorise: 2...
Question
Factorise:
2
√
2
a
3
+
16
√
2
b
3
+
c
3
−
12
a
b
c
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Solution
We know the identity
a
3
+
b
3
+
c
3
−
3
a
b
c
=
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
−
a
b
−
b
c
−
c
a
)
Using the above identity taking
a
=
√
2
a
,
b
=
2
√
2
b
and
c
=
c
, the equation
can be factorised as follows:
2
√
2
a
3
+
16
√
2
b
3
+
c
3
−
12
a
b
c
=
(
√
2
a
)
3
+
(
2
√
2
b
)
3
+
(
c
)
3
−
3
(
√
2
a
)
(
2
√
2
b
)
(
c
)
=
(
√
2
a
+
2
√
2
b
+
c
)
[
(
√
2
a
)
2
+
(
2
√
2
b
)
2
+
(
c
)
2
−
(
√
2
a
×
2
√
2
b
)
−
(
2
√
2
b
×
c
)
−
(
c
×
√
2
a
)
]
=
(
√
2
a
+
2
√
2
b
+
c
)
(
2
a
2
+
8
b
2
+
c
2
−
4
a
b
+
2
√
2
b
c
−
√
2
c
a
)
Hence,
2
√
2
a
3
+
16
√
2
b
3
+
c
3
−
12
a
b
c
=
(
√
2
a
+
2
√
2
b
+
c
)
(
2
a
2
+
8
b
2
+
c
2
−
4
a
b
+
2
√
2
b
c
−
√
2
c
a
)
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1
Similar questions
Q.
Factorize :
2
√
2
a
3
+
16
√
2
b
3
+
c
3
−
12
a
b
c
.
Q.
factorize: 2√2a
3
+16√2b
3
+c
3
-12abc
Q.
Factorise:
a
3
−
2
√
2
b
3
Q.
If 2A3 = 6, 2B3 = 5 and 2C3 = –1, then value of 5A5B5C5 is
यदि 2A3 = 6, 2B3 = 5 तथा 2C3 = –1, तब 5A5B5C5 का मान है
Q.
Factorise:
36
a
2
+
12
a
b
c
−
15
b
2
c
2
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