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Yosef115 wrote:
(A+B+C)² = ?
[A² + B² + C² +2(AB + BC + AC)]
The Riemann hypothesis states that when the Riemann zeta function crosses zero (except for those zeros between -10 and 0), the real part of the complex number has to equal to 1/2. That little claim might not sound very important. But it is. And we may be just a teensy bit closer to solving it ~if you're interested then the source of this statement is livescience.com
Yosef115 wrote:
(A–B+C)² = A²+B²+ C²– 2(AB–BC–CA). And you were wrong josh.
This is wrong...
(A–B+C)² = A²+B²+ C²+ 2(–AB–BC+CA)
This is correct, right?
sid0049 wrote:
Yosef115 wrote:
(A–B+C)² = A²+B²+ C²– 2(AB–BC–CA). And you were wrong josh.
This is wrong...
(A–B+C)² = A²+B²+ C²+ 2(–AB–BC+CA)
This is correct, right?
Yup.
((A) + (-B) + (C)) * ((A) + (-B) + (C))
A² + B² + C² + -AB + -AB + AC + AC + -BC + -BC
A² + B² + C² + -2AB + 2AC + -2BC
A² + B² + C² + 2(-AB + AC + -BC)
(or A² + B² + C² - 2(AB - AC + BC), whichever you prefer.)
Yosef115 wrote:
sid0049, see in video 'Algebra basics' they'll show the sum which I said.
Then they are wrong...
Yosef115 wrote:
sid0049, see in video 'Algebra basics' they'll show the sum which I said.
Your answer to the first one is correct. Since the second problem is just the first with the sign on B flipped, you can actually figure out the second one's answer by taking the pieces of the first one's answer and flipping the sign for each B in a given part: the AB and BC become negative for having one B each (and then you factor out -1 to get the form you presented), and the B² part flips twice, so it stays positive.
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