If x is a positive real number and exponents are rational numbers,then simply the -
(1) +
(x(a+b))2 (x(b+c))2(x(c+a))2. divided by
{x(a)x(b)x(c)}4
Answers
Answer:
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Step-by-step explanation:
Solution
To find the value of
[\frac{x^{b} }{x^{c} } ]^{b+c -a} .[\frac{x^{c} }{x^{a} } ]^{c+a-b} .[ \frac{x^{a} }{x^{b} } ]^{a+b-c}[
x
c
x
b
]
b+c−a
.[
x
a
x
c
]
c+a−b
.[
x
b
x
a
]
a+b−c
Here , we are applying property of exponent
\frac{x^{a} }{x^{b} } = x^{a-b}
x
b
x
a
=x
a−b
Therefore on applying this we have
[{x^{b-c} }]^{b+c -a} .[{x^{c-a} } ]^{c+a-b} .[ {x^{a-b} ]^{a+b-c}
This can be simplified as
= [ x^{(b+c)(b-c)} .x^{-a(b-c)} ] .[ x^{(c+a)(c-a)} .x^{-b(c-a)} ] .[ x^{(a+b)(a-b)} .x^{-c(a-b)} ]=[x
(b+c)(b−c)
.x
−a(b−c)
] .[x
(c+a)(c−a)
.x
−b(c−a)
] .[x
(a+b)(a−b)
.x
−c(a−b)
]
Here we are applying a^2 - b^2 = {a-b}{a+b}a
2
− b
2
= a−ba+b
We have
= [ x^{(b^{2} -c^{2} )} .x^{-ab+ac} ] .[ x^{(c^{2} -a^{2} )} .x^{-bc+ab} ] .[ x^{(a^{2} -b^{2} )} .x^{-ac+bc} ]=[x
(b
2
−c
2
)
.x
−ab+ac
] .[x
(c
2
−a
2
)
.x
−bc+ab
] .[x
(a
2
−b
2
)
.x
−ac+bc
]
Again using properties of exponent we have
= [ x^{(b^{2} -c^{2} +c^{2} -a^{2} + a^{2} -b^{2} )} ] .[ x^{(-bc+ab-ab-ca+bc+ca)} ]=[x
(b
2
−c
2
+c
2
−a
2
+a
2
−b
2
)
].[ x
(−bc+ab−ab−ca+bc+ca)
]
= [ x^{0} ] .[ x^{0} ]=[x
0
].[ x
0
]
= 1 . 1= 1.1
= 1
Hence , finally we get 1