Answer: 1
where x is nonzero
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Explanation:
We'll use two rules here
- (a^b)^c = a^(b*c) ... multiply exponents
- a^b*a^c = a^(b+c) ... add exponents
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The portion [ x^(a-b) ]^(a+b) would turn into x^[ (a-b)(a+b) ] after using the first rule shown above. That turns into x^(a^2 - b^2) after using the difference of squares rule.
Similarly, the second portion turns into x^(b^2-c^2) and the third part becomes x^(c^2-a^2)
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After applying rule 1 to each of the three pieces, we will have 3 bases of x with the exponents of (a^2-b^2), (b^2-c^2) and (c^2-a^2)
Add up those exponents (using rule 2 above) and we get
(a^2-b^2)+(b^2-c^2)+(c^2-a^2)
a^2-b^2+b^2-c^2+c^2-a^2
(a^2-a^2) + (-b^2+b^2) + (-c^2+c^2)
0a^2 + 0b^2 + 0c^2
0+0+0
0
All three exponents add to 0. As long as x is nonzero, then x^0 = 1
