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Question Number 141822 by mathsuji last updated on 23/May/21

Commented by MJS_new last updated on 23/May/21

$$\mathrm{where}\mathrm{you}\mathrm{found}\mathrm{this}?$$$$\mathrm{did}\mathrm{your}\mathrm{teacher}\mathrm{gave}\mathrm{it}\mathrm{to}\mathrm{you}?\mathrm{which}\mathrm{chapter}$$$$\mathrm{of}\mathrm{solving}\mathrm{equations}\mathrm{are}\mathrm{you}\mathrm{studying}?$$$$\mathrm{have}\mathrm{you}\mathrm{got}\mathrm{any}\mathrm{idea}?$$$$\bullet \mathrm{yes}:\mathrm{then}\mathrm{please}\mathrm{tell}\mathrm{us}$$$$\bullet \mathrm{no}:\mathrm{then}\mathrm{believe}\mathrm{me},{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{general}$$$$\mathrm{method}\mathrm{to}\mathrm{solve}\mathrm{it}.$$$$\mathrm{I}\mathrm{can}\mathrm{see}z=0\mathrm{solves}\mathrm{it}.\mathrm{can}\mathrm{you}\mathrm{see}\mathrm{anything}$$$$\mathrm{else}?$$

Commented by mathdanisur last updated on 23/May/21

$$1-z⩾0,1+x⩾0\Rightarrow -1⩽x⩽1$$$$a=\sqrt[20]{1-x},b=\sqrt[20]{1+x},a;b⩾0$$$$5(a^4+b^4)=2+4(a^5+b^5)$$$$4b^5-5b^4+1=-(4a^5-5a^4+1)$$$${(b-1)}^2(4b^3+3b^2+2b+1)=-{(a-1)}^2(4a^3+3a^2+2a+1)\left(i\right)$$$$fromLHSofequation\left(i\right)weobtain:$$$${(b-1)}^2(4b^3+3b^2+2b+1)⩾0$$$$andfromLHSofequation\left(i\right)$$$$-{(a-1)}^2(4a^3+3a^2+2a+1)⩽0$$$$sowecondindethat$$$$a-1=b-1$$$$a=b=2-x=1+x\Rightarrow x=0$$

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