New SIDBI Grade A New Batch Starts on Nov 18

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    Question

    A 1 Correct Answer Incorrect Answer
    B 2 Correct Answer Incorrect Answer
    C 3 Correct Answer Incorrect Answer
    D 4 Correct Answer Incorrect Answer

    Solution

    As we know, (x3 + y3) = (x + y) (x2 + y2 - xy) (x3 – y3) = (x - y) (x2 + y2 + xy) (x + y)2 - (x - y)2 = 4xy [(x - y)2 + 3xy] = [x2 + y2 - 2xy + 3xy] = [x2 + y2 + xy] [(x + y)2 - 3xy] = [x2 + y2 + 2xy - 3xy] = [x2 + y2 - xy] Now- (x3-y3)/x[(x+y)2-3xy] ÷y(x-y)2+3xy]/ x3+y3 ×(x+y)2-(x-y)2 /x2- y2 = (x - y) (x2 + y2 + xy)/x [x2 + y2 - xy] ÷ y [x2 + y2 + xy]/ (x + y) (x2 + y2 - xy) × 4xy/(x-y) (x+y) = (x - y) (x2 + y2 + xy)/x [x2 + y2 - xy] × (x + y) (x2 + y2 - xy)/ y [x2 + y2 + xy] × 4xy/(x-y) (x+y) =4

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