Radical Math Hacks Every Student Needs To Know!

[@ManojkantSamal]

Ans:(*13 -1) ^=read as to the power *=read as square root Let R=*32/(*26+*18) A=*162/(*26+*8) As per question (R+A)^(1/3)=? Let's explain R R=*32/(*26+*18) =(*16×*2)/{(*13×*2)+(*9×*2)} =(4.*2)/{*2(*13+3)} =4/(*13+3) ={4(*13-3)}/{(*13+3)(*13-3)} ={4(*13-3)}/4 =*13-3 Let's explain A A=*162/(*26+*8) =(*81×*2)/{(*13×*2)+(*4×*2)} =(9.*2)/{*2(*13+2)} =9/(*13+2) ={9(*13-2)}/{(*13+2)(*13-2)} ={9(*13-2)}/(13-4) ={9(*13-2)}/9 =*13-2 So, R+A=*13-3+*13-2 =(2.*13)-5 ={8(2.*13 -5)}/8 =(16.*13 -40)/8 ={13.*13+3.*13 -39-1}/8 ={13.*13-1-39-3.*13}/8 =[(*13)^3-(1^3)-{3×(*13^2)×1}+{3×**13×(1^2)}]/(2^3) =(*13-1)^3/(2^3) ={(*13-1)/2}^3 As per question (R+A)^(1/3) =[{(*13-1)/2}^3]^(1/3) =(*13 -1)/2

[@RashmiRay-c1y]

E=[2√13 -5]^1/3. E^3= 2√13 -5 and F^3=0.171 °C³ 2√13 +5. So, E^2-F^2=-10 and (EF)^3=27 and hence EF=3. Let E-F=t. Then, -10 = t[t^2+9]. So, t= E-F=-1 and EF=3. Thus, E=1/2[√3 -1].

[@潘博宇-k4l]

E=[2√13-5)^(1/3)=(√13-1)/2.

[@Quest3669]

X^3= 2√13-5 or 8x^3= 16√13-40 or x=( √13-1)/2 soln.

[@gregevgeni1864]

(√13 - 1)/2

[@Shobhamaths]

E=(√13-1) /2👍 ³√(√13-3+√13-2) ³√(2√13-5) x^3=2√13-5 y^3=2√13+5 x^3-y^3=-10 xy=3 x-y=-1 x^2+x-3=0

[@Fjfurufjdfjd]

[(13)^(1/2)-1]/2

[@petertang2524]

=（√13-1）/ 2