Question

if x= 2 + cube root(2) + cube root(4)
then find cube(x) - 6*square(x) + 6*x - 2

Answer 1

Answer:

Step-by-step explanation:

$(x-2)^3=(\sqrt[03]{2}+ \sqrt[3]{4})^3\\x^3-8-6x^2+12x=2+4+3*\sqrt[03]{16}+3*\sqrt[03]{32}\\\x^3-2-6x^2+6x+6x-6=6+6*\sqrt[03]{2}+6*\sqrt[03]{4}\\x^3-2-6x^2+6x=12+6*\sqrt[03]{2}+6*\sqrt[03]{4}-6x\\x^3-2-6x^2+6x=6*(2+\sqrt[03]{2}+\sqrt[03]{4})-6x\\\\substituting x=2+ cube root(2) + cube root(4)\\we get\\x^3-2-6x^2+6x=0$

Answer 2

Answer:

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Step-by-step explanation:

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