KeerthanaPosted on का सरलीकृत मूल्य 13⋅13⋅13+14⋅14⋅14+15⋅15⋅15−3⋅13⋅14⋅1513⋅13+14⋅14+15⋅15−(13⋅14+14⋅15+15⋅13)\frac{\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3}+\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5}-3 \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{5}}{\frac{1}{3} \cdot \frac{1}{3}+\frac{1}{4} \cdot \frac{1}{4}+\frac{1}{5} \cdot \frac{1}{5}-\left(\frac{1}{3} \cdot \frac{1}{4}+\frac{1}{4} \cdot \frac{1}{5}+\frac{1}{5} \cdot \frac{1}{3}\right)}31⋅31+41⋅41+51⋅51−(31⋅41+41⋅51+51⋅31)31⋅31⋅31+41⋅41⋅41+51⋅51⋅51−3⋅31⋅41⋅51 is a5160\frac{51}{60}6051 b4760\frac{47}{60}6047 c1360\frac{13}{60}6013 d4960\frac{49}{60}6049 Answer : Option BExplanation : Let 13=a;14=b\frac{1}{3}=a ; \frac{1}{4}=b31=a;41=b and 15=c\frac{1}{5}=c51=c ∴\therefore∴ Expression =a3+b3+c3−3abca2+b2+c2−ab−bc−ac=\frac{a^{3}+b^{3}+c^{3}-3 a b c}{a^{2}+b^{2}+c^{2}-a b-b c-a c}=a2+b2+c2−ab−bc−aca3+b3+c3−3abc =(a+b+c)(a2+b2+c2−ab−bc−ac)a2+b2+c2−ab−bc−ac=\frac{(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-a c\right)}{a^{2}+b^{2}+c^{2}-a b-b c-a c}=a2+b2+c2−ab−bc−ac(a+b+c)(a2+b2+c2−ab−bc−ac) =a+b+c=13+14+15=a+b+c=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=a+b+c=31+41+51 =20+15+1260=4760=\frac{20+15+12}{60}=\frac{47}{60}=6020+15+12=6047 Rate This:NaN / 5 - 1 votesAdd comment
