(x^a/x^b)^a^2+ab+b^2
Answers
Answer:
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Step-by-step explanation:
(\frac{x^{a}}{x^{b}})^{a^{2}+ab+b^{2}}(
x
b
x
a
)
a
2
+ab+b
2
\times (\frac{x^{b}}{x^{c}})^{b^{2}+bc+c^{2}}×(
x
c
x
b
)
b
2
+bc+c
2
\times (\frac{x^{c}}{x^{a}})^{c^{2}+ca+a^{2}}×(
x
a
x
c
)
c
2
+ca+a
2
=(x^{a-b})^{(a^{2}+ab+b^{2})} \times(x
a−b
)
(a
2
+ab+b
2
)
×
(x^{b-c})^{(b^{2}+bc+c^{2})}(x
b−c
)
(b
2
+bc+c
2
)
\times (x^{c-a})^{(c^{2}+ca+a^{2})}×(x
c−a
)
(c
2
+ca+a
2
)
=x^{(a-b)(a^{2}+ab+b^{2})}=x
(a−b)(a
2
+ab+b
2
)
\times x^{(b-c)(b^{2}+bc+c^{2})}×x
(b−c)(b
2
+bc+c
2
)
\times x^{(c-a)(c^{2}+ca+a^{2})}×x
(c−a)(c
2
+ca+a
2
)
=x^{a^{3}-b^{3}}\times x^{b^{3}-c^{3}}\times x^{c^{3}-a^{3}}=x
a
3
−b
3
×x
b
3
−c
3
×x
c
3
−a
3
=x^{a^{3}-b^{3}+b^{3}-c^{3}+c^{3}-a^{3}}=x
a
3
−b
3
+b
3
−c
3
+c
3
−a
3
=x^{0}=x
0
= 1=1
Therefore,
(\frac{x^{a}}{x^{b}})^{a^{2}+ab+b^{2}}(
x
b
x
a
)
a
2
+ab+b
2
\times (\frac{x^{b}}{x^{c}})^{b^{2}+bc+c^{2}} \times (\frac{x^{c}}{x^{a}})^{c^{2}+ca+a^{2}}×(
x
c
x
b
)
b
2
+bc+c
2
×(
x
a
x
c
)
c
2
+ca+a
2
= 1
