Suggested: x3+8=0 - x^3-8=0 - show that the straight lines x-2y+3=0 and 6x+3y+8=0 are perpendicular - 2x+3y=8 x-2y+3=0 solve graphically - 1/(x-2)(x-4)+1/(x-4)(x-6)+1/(x-6)(x-8)+1/3=0 - y=x^3 y=8 x=0 about x=3 - x+2y=3 2x-3y+8=0 - solve x^6+7x^3-8=0 - solve x3-7x2+14x-8=0 given that the roots are in geometric progression - x3+8=0 Browse related:
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