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Property 6

If x=3+3 1/...

Question

If $x = 3 + 3^{1 / 3} + 3^{2 / 3}$ then $x^{3} - 9 x^{2} + 18 x =$

A

0

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B

12

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C

- 12

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D

- 24

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Solution

The correct option is B $12$
question given
$x = 3 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}}$ - (1) find $x^{3} - 9 x^{2} + 18 x$
$(x - 3) = 3^{\frac{1}{3}} + 3^{\frac{2}{3}}$
therefore , cube both sides
$(x - 3)^{3} = (3^{\frac{1}{3}} + 3^{\frac{2}{3}})^{3}$ $(a - b)^{3} = a^{3} + b^{3} - 3 a b (a + b)$
$(a + b)^{3} = a^{3} + b^{3} + 3 a b (a + b)$
$x^{3} - 27 - 9 x^{2} + 27 x = 3 + 9 + 3 \times 3 (3^{\frac{1}{3}} + 3^{\frac{2}{3}})$
put (x-3)= $3^{\frac{1}{3}} + 3^{\frac{2}{3}}$
$x^{3} - 27 - 9 x^{2} + 27 x = 12 + 9 (x - 3)$
$x^{3} - 27 - 9 x^{2} + 27 x = 12 + 9 x - 27$
$x^{3} - 27 + 9 x^{2} + 27 x - 12 - 9 x + 27 = 0$
$x^{3} - 9 x^{2} + 18 x - 12 = 0$
$x^{3} - 9 x^{2} + 18 x = 12$

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Similar questions

x = 3 + 3^{1 / 3} + 3^{2 / 3}

x^{3} - 9 x^{2} + 18 x - 12

=

x = 3 + 3^{\frac{1}{2}} + 3^{\frac{2}{3}}

x^{3} - 9 x^{2} + 18 x - 12

x = 3 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}}

, then find the value of

x^{3} - 9 x^{2} + 18 x

If [x=3+3^(1/3)+3^(2/3)] , then prove that [(x^3)-(9x^2)+18x-12]=0

y = x^{4} + x^{3} - 18 x^{2} + 24 x - 12

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