Question 58

If $$x=\frac{3}{2}$$, then the value of $$27x^3-54x^2+36x-11$$ is

Solution

Given : $$x=\frac{3}{2}$$

To find : $$27x^3-54x^2+36x-11$$

= $$27(\frac{3}{2})^3-54(\frac{3}{2})^2+36(\frac{3}{2})-11$$

= $$(27 \times \frac{27}{8})-(54 \times \frac{9}{4})+(54)-11$$

= $$\frac{729}{8}-\frac{486}{4}+43$$

= $$\frac{729-972+344}{8}=\frac{101}{8}$$

= $$12\frac{5}{8}$$

=> Ans - (D)

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