Question

1 1 if x3 + 1 = 110, then x + = 23 T O 15 O 5 O 10 O 25​

Answer 1

Answer:

(x+1)=3x resolves with easiness as: 2x=1=2×.5

x2+1=3x however, we observe this does not atall result same easiness of resolve as 1, and we get to Figure 3 as: (Example1×(x OR 1)) To share that Same Resolve as Figure 1 must also equal 3x(y OR 1) with something more alike 2xy+x=xy+2x=3xy=x2+1 via y=1 onsets that a Factor is thereat via (y or 1)=y=1 has Equality of the Two Options. And we know .5 cannot equal 1, so we know x is not equal 1 nor .5 within status of Parenthetical Shift (See 8 below)

(x−(−1))(x+(−1))=3x=x2−(−1) is the Breakdown of this obvious permutation of Figure 1, which truly would have more easily Formulated: 3x−2=x2−1=(x−1)(x+1) atwhich we know LHS<RHS at x>1 and x<2 LHS<RHS at x<1 we need to thus by Isoscelean Surveyance more approach a Box, as Thus is 3x2−x=x3 is Standing Near enough to recognize the x3 as a Box. And the 3x2−x is Three Imperfect Spheres likely, forensically associable a Former state that: These were 3x2 when inside the x3 Box, but the Contents became Scattered and Spoiled or Peeled.