If α=45+29√23+45-29√23, then the value of ∝3-21∝-90 _______.

According to the problem, we are given α=45+29√23+45-29√23 and we need to find the value of ∝3-21∝-90 .Let us assume α= x13+y13 such that x=45+29√2, y=45-29√2.Let us first find the value of ∝3.So, we have α=x13+y13Let us now apply cubing on both sides.⇒α3=(x13+y13)3We know that (a+b)3=a3+3a2b+3ab2+b3⇒α3=(x13)3+3(x13)2b+3a(y13)2+(y13)3⇒ α3=x+3(x13y13)x13+3(x13y13) y13+yOn simplifying, we get⇒ α3=x+3(xy)13x13+3(xy)13y13+yTaking factor (xy)13 common, we get⇒α3=(x+y)+3xy13(x13+y13)....(1)Now we will find the value of xy,⇒(xy)13 = ((45+29√2)(45-29√2))13⇒(xy)13=(2025-10352+10352-1682))13.On Simplifying, we get⇒(xy)13=(343)13As we know that,343=73So, (xy)13 =(73)13On solving we get⇒(xy)13=7 ---(2).Let us substitute equation (2) in equation (1).⇒α3=(x+y)+3(7)(x13+y13)But we already know that α=(x13+y13 such that x=45+29√2, y=45-29√2.⇒α3=(45+292+45-292) +21α.On solving, we get⇒α3 =90+21α⇒α3−21α−90=0∴ We have found the value of α3−21α−90=0 as 0.

If α=45+29√23+45-29√23, then the value of ∝3-21∝-90 _______.

According to the problem, we are given α=45+29√23+45-29√23 and we need to find the value of ∝3-21∝-90 .Let us assume α= x13+y13 such that x=45+29√2, y=45-29√2.Let us first find the value of ∝3.So, we have α=x13+y13Let us now apply cubing on both sides.⇒α3=(x13+y13)3We know that (a+b)3=a3+3a2b+3ab2+b3⇒α3=(x13)3+3(x13)2b+3a(y13)2+(y13)3⇒ α3=x+3(x13y13)x13+3(x13y13) y13+yOn simplifying, we get⇒ α3=x+3(xy)13x13+3(xy)13y13+yTaking factor (xy)13 common, we get⇒α3=(x+y)+3xy13(x13+y13)....(1)Now we will find the value of xy,⇒(xy)13 = ((45+29√2)(45-29√2))13⇒(xy)13=(2025-10352+10352-1682))13.On Simplifying, we get⇒(xy)13=(343)13As we know that,343=73So, (xy)13 =(73)13On solving we get⇒(xy)13=7 ---(2).Let us substitute equation (2) in equation (1).⇒α3=(x+y)+3(7)(x13+y13)But we already know that α=(x13+y13 such that x=45+29√2, y=45-29√2.⇒α3=(45+292+45-292) +21α.On solving, we get⇒α3 =90+21α⇒α3−21α−90=0∴ We have found the value of α3−21α−90=0 as 0.