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This solution makes very nice use of the fact that a technique can be repeated in the same problem -- here by applying difference of squares more than once, even with the expression 6 3 2 − 3 7 2 which some might stop at but lends itself to the simple expression ( 1 0 0 ) ( 2 6 ) .
Even when a calculation involves pure numbers, recognizing algebraic patterns can be useful!
Yup, the steps becomes abundantly clear once we've identified that it's just a (repeated) difference of squares.
In this case, the expression is in the form of (a^4-b^4)/(a^2 + b^2), the numerator factors to (a^2-b^2)(a^2 + b^2)
And we're left with (a^2 - b^2), which can be further broken down to (a-b)(a+b)
Where did I go wrong? Squaring everything once should leave me with 63^2 - 37^2 /63+37 -> 63^2 - 37^2 /100. Squaring everything again leaves me with 63 - 37/10 = 2.6
Shouldn't I multiply it by 10^4? It gives me a result of 26000. Where is my mistake?
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You can't just remove the squares like that. ( a 2 − b 2 ) / ( c 2 − d 2 ) is not equal to ( a − b ) / ( c − d ) .
we should make it easy 63 * 63=3609 * 63=10287+8997=5542 * 544/665 answer 2600
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I don't understand your sequences of equations.
The point of the question is that we don't have to evaluate 6 3 2 .
Nice use of colors :) We can use the difference of two squares identity once again for 6 3 2 − 3 7 2 to further simplify the calculations.
6 3 − 3 7 = 2 6 , and 6 3 + 3 7 = 1 0 0 . It is easy to multiply 26 with 100 to get 2600.
cool - how'd you make that image, please?
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Terry, we can use LaTeX on this site to render math
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Yes and I've done that, but this is a graphic image.
6 3 2 + 3 7 2 6 3 4 − 3 7 4 = 6 3 2 + 3 7 2 ( 6 3 2 − 3 7 2 ) ( 6 3 2 + 3 7 2 ) = 6 3 2 − 3 7 2 = ( 6 3 − 3 7 ) ( 6 3 + 3 7 ) = 2 6 ( 1 0 0 ) = 2 6 0 0
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6 3 2 + 3 7 2 6 3 4 − 3 7 4 = ( 6 3 2 + 3 7 2 ) ( 6 3 2 + 3 7 2 ) ( 6 3 2 − 3 7 2 ) = 6 3 2 − 3 7 2 = ( 6 3 + 3 7 ) ( 6 3 − 3 7 ) = ( 1 0 0 ) ( 2 6 ) = 2 6 0 0