Nice Algebra Problem | You should be able to solve this!

[@vijayannair2316]

Nice

[@Nature-Melody2106]

Amazing❤❤❤

[@dileepmv7438]

but he is using trial and error to find out that 7 is a factor; this is a wrong method

[@MathBooster]

Trial and error is the only method to solve cubic equation for up to grade 12 students (in higher studies, you will learn a complex formula to solve cubic equation). So, it is not wrong. Examinor will always give you only such cubic equations to solve in which you can easily find one solution by trial and error.

[@BRUBRUETNONO]

Hi, Thanks for your interesteing problem. Here is how I solved it without looking at your solution. I hope you like it. Greetings ! xy=3 x^6+y^6=154 x^8+y^8=? We have (x^2+y^2)^3=(x^2)^3+(y^2)^3+3x^2y^2(x^2+y^2) (x^2+y^2)^3=(x^6+y^6)+3(xy)^2(x^2+y^2) If we set q=x^2+y^2 then q^3=154+3.3^2q q^3-(343-7.27)-27q=0 (q^3-343)-27(q-7)=0 (q^3-7^3)-27(q-7)=0 (q-7)(q^2+7q+7^2)-27(q-7)=0 (q-7)(q^2+7q+7^2-27)=0 (q-7)(q^2+7q+22)=0 q-7=0 or q^2+7q+22=0 q^2+7q+22=0 has a negative discriminant as D=7^2-4.22