If $$x-y-\sqrt{18}=-1$$ and $$x + y - 3\sqrt{2} = 1$$, then what is the value of $$12xy(x^{2} - y^{2})$$ ?
Given : $$x-y-\sqrt{18}=-1$$
=> $$x-y=\sqrt{18}-1$$ -------------(i)
Squaring both sides,
=> $$(x-y)^2=(\sqrt{18}-1)^2$$
=> $$x^2+y^2-2xy=18+1-2\sqrt{18}$$
=> $$x^2+y^2-2xy=19-2\sqrt{18}$$ --------------(ii)
Also, $$x + y - 3\sqrt{2} = 1$$
=> $$x+y=\sqrt{18}+1$$ -------------(iii)
Squaring both sides,
=> $$(x+y)^2=(\sqrt{18}+1)^2$$
=> $$x^2+y^2+2xy=18+1+2\sqrt{18}$$
=> $$x^2+y^2+2xy=19+2\sqrt{18}$$ --------------(iv)
Subtracting equation (ii) from (iv),
=> $$4xy=4\sqrt{18}$$
=> $$12xy=12\sqrt{18}$$ ------------(v)
Multiplying equations (i) and (iii),Â
=> $$(x-y)(x+y)=(\sqrt{18}-1)(\sqrt{18}+1)$$
=> $$x^2-y^2=18-1=17$$ -----------(vi)
Now, multiplying equations (v) and (vi), we get :
=>Â $$12xy(x^{2} - y^{2})= (12\sqrt{18})\times17$$
=Â $$204\sqrt{18}=612\sqrt2$$
=> Ans - (D)
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