Question 155

If $$(m+1) = \sqrt{n}+3$$ the value of $$\frac{1}{2}(\frac{m^{3}-6m^{2}+12m-8}{\sqrt{n}}-n)$$

Solution

If $$(m+1) = \sqrt{n}+3$$

=> $$m-2 = \sqrt{n}$$ --------------Eqn(1)

to find : $$\frac{1}{2}(\frac{m^{3}-6m^{2}+12m-8}{\sqrt{n}}-n)$$

$$\because (m-2)^3 = m^{3}-6m^{2}+12m-8$$

=> $$\frac{1}{2}(\frac{(m-2)^3}{\sqrt{n}}-n)$$

Using eqn(1), we get :

=> $$\frac{1}{2}(\frac{(\sqrt{n})^3}{\sqrt{n}}-n)$$

=> $$\frac{1}{2}(n-n)$$

= 0

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