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Question Number 148522 by mathdanisur last updated on 28/Jul/21

Answered by Rasheed.Sindhi last updated on 29/Jul/21

P(x)=16x^2 +24x+9 Q(x)=x^3 −3x^2 +2x+2 If x=−1.25 ((P^3 +Q^3 )/(P^2 −PQ+Q^2 ))+((P^3 −Q^3 )/(P^2 +PQ+Q^2 ))=? ((P^3 +Q^3 )/(P^2 −PQ+Q^2 ))+((P^3 −Q^3 )/(P^2 +PQ+Q^2 )) =(P+Q)+(P−Q) =2P=2(16x^2 +24x+9) =2(16(−(5/4))^2 +24(−(5/4))+9) =2(16(((25)/(16)))−6(5)+9) =2(25−30+9)=2(4)=8

$${P}\left({x}\right)=\mathrm{16}{x}^{\mathrm{2}} +\mathrm{24}{x}+\mathrm{9} \\ $$$${Q}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2} \\ $$$${If}\:{x}=−\mathrm{1}.\mathrm{25} \\ $$$$\frac{{P}^{\mathrm{3}} +{Q}^{\mathrm{3}} }{{P}^{\mathrm{2}} −{PQ}+{Q}^{\mathrm{2}} }+\frac{{P}^{\mathrm{3}} −{Q}^{\mathrm{3}} }{{P}^{\mathrm{2}} +{PQ}+{Q}^{\mathrm{2}} }=?\: \\ $$$$\frac{{P}^{\mathrm{3}} +{Q}^{\mathrm{3}} }{{P}^{\mathrm{2}} −{PQ}+{Q}^{\mathrm{2}} }+\frac{{P}^{\mathrm{3}} −{Q}^{\mathrm{3}} }{{P}^{\mathrm{2}} +{PQ}+{Q}^{\mathrm{2}} } \\ $$$$=\left({P}+{Q}\right)+\left({P}−{Q}\right) \\ $$$$=\mathrm{2}{P}=\mathrm{2}\left(\mathrm{16}{x}^{\mathrm{2}} +\mathrm{24}{x}+\mathrm{9}\right) \\ $$$$\:\:\:\:\:=\mathrm{2}\left(\mathrm{16}\left(−\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} +\mathrm{24}\left(−\frac{\mathrm{5}}{\mathrm{4}}\right)+\mathrm{9}\right) \\ $$$$\:\:\:\:\:=\mathrm{2}\left(\mathrm{16}\left(\frac{\mathrm{25}}{\mathrm{16}}\right)−\mathrm{6}\left(\mathrm{5}\right)+\mathrm{9}\right) \\ $$$$\:\:\:\:\:=\mathrm{2}\left(\mathrm{25}−\mathrm{30}+\mathrm{9}\right)=\mathrm{2}\left(\mathrm{4}\right)=\mathrm{8} \\ $$$$ \\ $$

Commented by mathdanisur last updated on 28/Jul/21

Cool Ser, thank you

$${Cool}\:{Ser},\:{thank}\:{you} \\ $$

Answered by Olaf_Thorendsen last updated on 28/Jul/21

R = ((P^3 +Q^3 )/(P^3 −PQ+Q^2 ))+((P^3 −Q^3 )/(P^3 +PQ+Q^2 )) R = (((P+Q)(P^2 −PQ+Q^2 ))/(P^3 −PQ+Q^2 ))+(((P−Q)(P^2 +PQ+Q^2 ))/(P^3 +PQ+Q^2 )) R = P+Q+P−Q = 2P R(−(5/4)) = 2(16.((25)/(16))−24.(5/4)+9) R(−(5/4)) = 8

$$\mathrm{R}\:=\:\frac{{P}^{\mathrm{3}} +{Q}^{\mathrm{3}} }{{P}^{\mathrm{3}} −{PQ}+{Q}^{\mathrm{2}} }+\frac{{P}^{\mathrm{3}} −{Q}^{\mathrm{3}} }{{P}^{\mathrm{3}} +{PQ}+{Q}^{\mathrm{2}} } \\ $$$$\mathrm{R}\:=\:\frac{\left({P}+{Q}\right)\left({P}^{\mathrm{2}} −{PQ}+{Q}^{\mathrm{2}} \right)}{{P}^{\mathrm{3}} −{PQ}+{Q}^{\mathrm{2}} }+\frac{\left({P}−{Q}\right)\left({P}^{\mathrm{2}} +{PQ}+{Q}^{\mathrm{2}} \right)}{{P}^{\mathrm{3}} +{PQ}+{Q}^{\mathrm{2}} } \\ $$$$\mathrm{R}\:=\:{P}+{Q}+{P}−{Q}\:=\:\mathrm{2}{P} \\ $$$${R}\left(−\frac{\mathrm{5}}{\mathrm{4}}\right)\:=\:\mathrm{2}\left(\mathrm{16}.\frac{\mathrm{25}}{\mathrm{16}}−\mathrm{24}.\frac{\mathrm{5}}{\mathrm{4}}+\mathrm{9}\right) \\ $$$${R}\left(−\frac{\mathrm{5}}{\mathrm{4}}\right)\:=\:\mathrm{8} \\ $$

Commented by mathdanisur last updated on 28/Jul/21

Cool Ser, thank you

$${Cool}\:{Ser},\:{thank}\:{you} \\ $$

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