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Step by Step Solution

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

7*(x+8)^6*(x-7)^3+(x+8)^7-(3*(x-7)^2)=0

Step 1 :

Equation at the end of step 1 :

  (((7•((x+8)⁶))•((x-7)³))+((x+8)⁷))-3•(x-7)²  = 0

Step 2 :

Equation at the end of step 2 :

  ((7•(x+8)⁶•(x-7)³)+(x+8)⁷)-3•(x-7)²  = 0

Step 3 :

Equation at the end of step 3 :

  (7•(x+8)⁶•(x-7)³+(x+8)⁷)-3•(x-7)²  = 0 
 Step  4  :
 4.1     Evaluate :  (x+8)⁶   =    x⁶+48x⁵+960x⁴+10240x³+61440x²+196608x+262144 
 4.2     Evaluate :  (x-7)²   =    x²-14x+49

Equation at the end of step 4 :

  7x⁹ + 189x⁸ + 692x⁷ - 22505x⁶ - 203952x⁵ + 558656x⁴ + 11425792x³ + 15568893x² - 204144598x - 631505043  = 0

Step 5 :

Equations of order 5 or higher :

5.1 Solve 7x⁹+189x⁸+692x⁷-22505x⁶-203952x⁵+558656x⁴+11425792x³+15568893x²-204144598x-631505043 = 0

Points regarding equations of degree five or higher.

(1) There is no general method (Formula) for solving polynomial equations of degree five or higher.

(2) By the Fundamental theorem of Algebra, if we allow complex numbers, an equation of degree n will have exactly n solutions
(This is if we count double solutions as 2 , triple solutions as 3 and so on

) (3) By the Abel-Ruffini theorem, the solutions can not always be presented in the conventional way using only a finite amount of additions, subtractions, multiplications, divisions or root extractions

(4) If F(x) is a polynomial of odd degree with real coefficients, then the equation F(X)=0 has at least one real solution.

(5) Using methods such as the Bisection Method, real solutions can be approximated to any desired degree of accuracy. Failed to find the initial interval for implementing the BiSection Method