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Byju's Answer
Standard IX
Mathematics
Factor of Polynomials
Find the prod...
Question
Find the product using appropriate identity:
(
a
−
b
−
c
)
(
a
2
+
b
2
+
c
2
+
a
b
−
b
c
+
c
a
)
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Solution
We know the identity
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
−
a
b
−
b
c
−
c
a
)
=
a
3
+
b
3
+
c
3
−
3
a
b
c
Using the above identity taking
a
=
a
,
b
=
−
b
and
c
=
−
c
, the product
(
a
−
b
−
c
)
(
a
2
+
b
2
+
c
2
+
a
b
−
b
c
+
c
a
)
can be computed as follows:
(
a
−
b
−
c
)
(
a
2
+
b
2
+
c
2
+
a
b
−
b
c
+
c
a
)
=
a
3
+
(
−
b
)
3
+
(
−
c
)
3
−
(
3
a
×
−
b
×
−
c
)
=
a
3
−
b
3
−
c
3
−
3
a
b
c
Hence,
(
a
−
b
−
c
)
(
a
2
+
b
2
+
c
2
+
a
b
−
b
c
+
c
a
)
=
a
3
−
b
3
−
c
3
−
3
a
b
c
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1
Similar questions
Q.
The expression
a
3
+
b
3
+
c
3
−
3
a
b
c
can be expressed as a product of two expressions. What is the product?
Q.
Find the product of
(
a
−
b
−
c
)
(
a
2
+
b
2
+
c
2
−
a
b
+
b
c
−
c
a
)
without actual multiplication
Q.
Find the product (a
− b − c
) (a
2
+ b
2
+ c
2
+ ab + ac
−
b
c).
Q.
Show that the determinant
∣
∣ ∣ ∣
∣
a
2
+
b
2
+
c
2
b
c
+
c
a
+
a
b
b
c
+
c
a
+
a
b
b
c
+
c
a
+
a
b
a
2
+
b
2
+
c
2
b
c
+
c
a
+
a
b
b
c
+
c
a
+
a
b
b
c
+
c
a
+
a
b
a
2
+
b
2
+
c
2
∣
∣ ∣ ∣
∣
is always non-negative.
Q.
If
∣
∣ ∣
∣
a
b
c
b
c
a
c
a
b
∣
∣ ∣
∣
=
k
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
−
b
c
−
c
a
−
a
b
)
, then
k
=
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