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B
y(ax−by−cz)
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C
xy(ax+by+cz)
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D
xy(a+by+cz)
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Solution
The correct option is Dxy(ax+by+cz) We have, ax2y=a×x×x×y bxy2=b×x×y×y and, cxyz=c×x×y×z The three terms have x and y as common factors ∴ax2y+bxy2+cxyz=(a×x×x×y)+(b×x×y×y)+(c×x×y×z) =x×y×(a×x+b×y+c×z) =xy(ax+by+cz)