1975 AHSME Problems/Problem 12

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We can factor $a^3-b^3=19x^3$ into: \[(a-b)(a^2+ab+b^2)=19x^3.\] Substituting yields: \[x(a^2+ab+b^2)=19x^3\] \[a^2+ab+b^2=19x^2.\] This is equal to: \[(a-b)^2+3ab=19x^2\] \[x^2+3ab=19x^2\] \[ab=6x^2.\] Checking with the possible answers, along with $a-b=x$ yields the only answer to be \box{(B): $a=3x$ or $a=-2x$}.