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Byju's Answer
Standard VI
Mathematics
Solving Equations x+a = b and x-a = b by Rearranging Terms
x-2 y3+2 y-3 ...
Question
(x – 2y)
3
+ (2y – 3z)
3
+ (3z – x)
3
Open in App
Solution
To
solve
:
x
-
2
y
3
+
2
y
-
3
z
3
+
3
z
-
x
3
Let
a
=
x
-
2
y
,
b
=
2
y
-
3
z
,
c
=
3
z
-
x
.
a
+
b
+
c
=
x
-
2
y
+
2
y
-
3
z
+
3
z
-
x
=
x
-
2
y
+
2
y
-
3
z
+
3
z
-
x
=
0
As
we
know
that
,
if
a
+
b
+
c
=
0
,
then
a
3
+
b
3
+
c
3
=
3
a
b
c
Therefore
,
a
3
+
b
3
+
c
3
=
3
a
b
c
⇒
x
-
2
y
3
+
2
y
-
3
z
3
+
3
z
-
x
3
=
3
x
-
2
y
2
y
-
3
z
3
z
-
x
Hence
,
x
-
2
y
3
+
2
y
-
3
z
3
+
3
z
-
x
3
=
3
x
-
2
y
2
y
-
3
z
3
z
-
x
Suggest Corrections
2
Similar questions
Q.
(x - 2y)
3
+ (2y - 3z)
3
+ (3z -x)
3
= .......
Q.
Simplify:
(
x
−
2
y
)
3
+
(
2
y
−
3
z
)
3
+
(
3
z
−
x
)
3
Q.
Factorize:
(
x
−
2
y
)
3
+
(
2
y
−
3
z
)
3
+
(
3
z
−
x
)
3
.
Q.
Without finding the cubes, factorise
(
x
−
2
y
)
3
+
(
2
y
−
3
z
)
3
+
(
3
z
−
x
)
3
Q.
Question 38
Without finding the cubes, factorise:
(
x
−
2
y
)
3
+
(
2
y
−
3
z
)
3
+
(
3
z
−
x
)
3
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Standard VI Mathematics
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