Show that xa - b2xb - c2xc - a2xaxbxc4- 1

( #10;x ^ a+b ) #160; ^ 2 ( x ^ b+c ) #160; ^ 2 #10;( x ^ a+c ) #160; ^ 2 ( x ^ a x ^ b #10; x ^ c ) #160; ^ 4 = x ^ 2a+2b+2b+2c+ ...

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Byju's Answer

Standard VIII

Mathematics

Expansion of (x^2 - y^2)

Show that x...

Question

Show that
$\frac{{(x^{a + b})}^{2} {(x^{b + c})}^{2} {(x^{c + a})}^{2}}{{(x^{a} x^{b} x^{c})}^{4}} = 1$

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Solution

$\frac{{(x^{a + b})}^{2} {(x^{b + c})}^{2} {(x^{a + c})}^{2}}{{(x^{a} x^{b} x^{c})}^{4}} = \frac{x^{2 a + 2 b + 2 b + 2 c + 2 a + 2 c}}{x^{4 a + 4 b + 4 c}} = x^{4 a + 4 b + 4 c - 4 a - 4 b - 4 c} = x^{0} = 1$

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Similar questions

Q. The value of

\frac{{(x^{a + b})}^{2} {(x^{b + c})}^{2} {(x^{c + a})}^{2}}{{(x^{a} x^{b} x^{c})}^{4}}

is equal to:

Prove that :

(i) ${(\frac{x^{a}}{x^{b}})}^{a^{2} + a b + b^{2}} \times {(\frac{x^{b}}{x^{c}})}^{b^{2} + a b + c^{2}} \times {(\frac{x^{c}}{x^{a}})}^{c^{2} + c a + a^{2}} = 1$

(ii) ${(\frac{x^{a}}{x^{b}})}^{c} \times {(\frac{x^{b}}{x^{c}})}^{a} \times {(\frac{x^{c}}{x^{a}})}^{b} = 1$

Q. The value of

{(\frac{x^{a}}{x^{b}})}^{a^{2} + a b + b^{2}} \cdot {(\frac{x^{b}}{x^{c}})}^{b^{2} + b c + c^{2}} \cdot {(\frac{x^{c}}{x^{a}})}^{c^{2} + c a + b^{2}}

Q. Simplify :

{(\frac{x^{a}}{x^{b}})}^{a^{2} + a b + b^{2}} \times {(\frac{x^{b}}{x^{c}})}^{b^{2} + b c + c^{2}} \times {(\frac{x^{c}}{x^{a}})}^{c^{2} + c a + a^{2}}

Q. If

∣ ∣ ∣ ∣ \begin{matrix} a + x & b + x & c + x b + x & c + x & a + x c + x & a + x & b + x \end{matrix} ∣ ∣ ∣ ∣ = 0

then

x =

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Show that (xa+b)2(xb+c)2(xc+a)2(xaxbxc)4=1

(xa+b)2(xb+c)2(xa+c)2(xaxbxc)4=x2a+2b+2b+2c+2a+2cx4a+4b+4c=x4a+4b+4c−4a−4b−4c=x0=1

Show that
$\frac{{(x^{a + b})}^{2} {(x^{b + c})}^{2} {(x^{c + a})}^{2}}{{(x^{a} x^{b} x^{c})}^{4}} = 1$

$\frac{{(x^{a + b})}^{2} {(x^{b + c})}^{2} {(x^{a + c})}^{2}}{{(x^{a} x^{b} x^{c})}^{4}} = \frac{x^{2 a + 2 b + 2 b + 2 c + 2 a + 2 c}}{x^{4 a + 4 b + 4 c}} = x^{4 a + 4 b + 4 c - 4 a - 4 b - 4 c} = x^{0} = 1$