Solve the equation $$8x-3+\sqrt{x+2}-\sqrt{x-1}=7 \sqrt{x^2+x-2}$$
I have this idea: set $$\sqrt{x+2}=a , x+2=a^2 , \sqrt{x-1}=b.$$ So $$x-1=b^2 , 2a^2+6b^2 =8b-4$$ and $$x^2+x-2 =a^2b^2$$ and then I'd simplify, but it's still very hard to solve. Any hint is appreciated.
The domain gives $x\geq1$ and we need to solve that $$\sqrt{x+2}-\sqrt{x-1}=7\sqrt{x^2+x-2}-8x+3.$$ Now, since $\sqrt{x+2}-\sqrt{x-1}\geq0,$ we obtain $$7\sqrt{x^2+x-2}-8x+3\geq0$$ or $$\frac{97-\sqrt{2989}}{30}\leq x\leq\frac{97+\sqrt{2989}}{30}.$$ Thus, we need to solve $$\left(\sqrt{x+2}-\sqrt{x-1}\right)^2=\left(7\sqrt{x^2+x-2}-(8x-3)\right)^2$$ or $$(112x-44)\sqrt{x^2+x-2}=113x^2-x-90$$ or $$(112x-44)^2(x^2+x-2)=(113x^2-x-90)^2$$ or $$225x^4-2914x^3+12669x^2-21468x+11972=0$$ or $$(25x-146)(x-2)(9x^2-46x+41)=0,$$ which gives the answer: $$\left\{2,\frac{23+4\sqrt{10}}{9}\right\}.$$ Also, your idea helps.
Indeed, we got the following system: $$a^2-b^2=3$$ and $$8(a^2-2)-3+a-b=7ab.$$ From the second equation we obtain: $$b=\frac{8a^2+a-19}{7a+1},$$ which gives $$a^2-\left(\frac{8a^2+a-19}{7a+1}\right)^2=3$$ or $$(5a+14)(a-2)(3a^2-2a-13)=0$$ and the rest is smooth.