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Amal Kumar sharing with you a test question on polynomial equations well solve this equation in the domain of complex numbers the question is find roots of the equation X to the power of 6 minus 26 X cubed minus 27 equals to 0 where X belongs to complex numbers so let me rewrite this equation it is X to the power of 6 minus 26 X cubed minus 27 equals to 0 now since we are working in the domain of complex number combined roots real and complex I mean all the roots should be since the degree is 6 we are expecting 6 roots remember that part right and now lets begin to solve this equation we can make a substitution substitution could be let X cube equals to P in that case we get a quadratic equation P squared minus 26 P minus 27 equals to 0 right so I hope that is clear if X cube is P then X cube square which is X to power of 6 will be P squared so we substituted this value now you could factor minus 26 is the sum product is 27 so we could write P minus 27 times P plus 1 equals to 0 so t
