hello and welcome now to solve this m olyp problem x = 3 + 8 and that x^ 6 - M XB + 1 = 0 what's the value of M we would like to find the value of this m here solution now from this first equation here we have X = to squ root of 3 +un of 8 now let us change the form of8 8 is the same as 4 * 2 so we have squ root of 3 +un of 4 * 2 now here apply this rule that squ of a * p is the same Asun of a * root of P now next we have this one equals to square root of 3 +un 4 * 2 now the of 4 is 2 so we have X = < TK of 3 + 2 < tk2 now this one can be F written as X = to S root of 3 + 2 * 1 * of 2 now this one let us write three in another form that 3 is the same as 1 + 2 so we have X = to root of 1 + 2 + 2 * 1 * < TK 2 now here 1 + 2 can be written as 1 2 +un of 2 squ now we have X = root of 1 + < TK 2 2 + 2 * 1 * < TK 2 now here this one forms the complete Square so let a be equal to 1 and P equals to S root of 2 so that a 2 + p 2 + 2 * a * P = to a + p² Now by use of this identity has change the form of this so we have X = to square root of in this form now we have 1 + square root of 2 of this Square that so that this one cancels with this we have x = 1+ s of 2 recall X to^ 6us MX cubed + 1 = 0 now let us put the value of x here so it implies that 1 + squ root of 2 of this ra to^ 6 - M into 1 + 2 of this to^ 3 + 1 = 0 now here let us make M the subject so here we have M into 1 + of 2 cubed = to 1 + of 2 of this to^ 6 + 1 now here divide both sides Pi divide by 1 + sare root of 2 of this ra to^ 3 so we have M = to 1 + 2 of this ra to^ 6 over 1 +un of 2 of this ra power 3 + 1 / 1 + 2 all of this ra to^ 3 now here as you can see the base is the same now apply this rule that a^ m/ a^ n is the same as a^ M - n so let us apply this rule here so we have M = to 1 +un of 2 ra^ 6 - 3 + 1 / 1 + < TK 2 of this is^ 3 next we have M = to 1 + 2 of this to^ 3 + 1 / 1 + of 2 cubed now let us apply this rule here that 1 / a to^ n is the same as 1 / a the of this ra to power n so then this Ru is going to be applied here so we have M = to 1 + < TK 2 cubed + 1 / 1 + of two the of this cubed now let us rationalize the denominate of this so it means that 1 / 1 + of 2 is the same as 1 / 1 + 2 * 1 - 2 over 1 - of 2 which gives us 1 - 2 over 1 squared - 2 which gives us s < TK of 2 - 1 over one now let us substitute this one here now next we have m equal to 1 + s of 2 cubed plus now we put this new form we have square of 2 - 1 of this cubed here let 1 + of 2 = to Y and of 2 - 1 = to Z so that y cubed + + z cubed is the same as y + z into y^ 2 - y z + c^ 2 the soal the sum of two cubes Now by use of this identity so we have M = to 1 + + 2 +un 2 - 1 into 1 +un of 2 2us into of 2 + + 1 * s < TK of 2 - 1 + squ < TK of 2 - 1 the of this squared next by expansion we have m equals two now this one here as you can see this one cancels with this so we have 2 < tk2 into this one becomes by expansion we have one + 2 < tk2 + 2 minus now this one forms the difference of the squares so this one and this one gives us one minus one from this two plus now this one gives us 2 - 2 < tk2 + 1 like this next m equals to 2 < tk2 into as you can see here this one cancels with this so we have 2 < tk2 and this 2 root2 cancels because of the negative sign so this one and this one cancels so we have 2 + 2 + 1 so we with 2 + 2 + 1 now this one gives us m equals to 2 ot2 times this one gives us five so finally we have M = to 10 < tk2 and that is it thank you for watching subscribe to my channel and turn on the notification Bell to get new updates when I up new videos don't forget to smash like button share and comment below to make this video many people for