Question

If $a - b = 3$and ${a^3} - {b^3} = 117$, then find the value of $a + b$

Answer 1

Hint:First, we need to know about the concept of algebraic identity in algebra.
These are the identities that will always be true for every value of variables in them and the applications in the factorization of the polynomial with the variables and constants.
In algebraic identity, the left-hand side equation always equals the right-hand side equation.

Formula used:
${a^3} - {b^3} = (a - b)({a^2} + {b^2} + ab)$
${(a + b)^2} = {(a - b)^2} + 4ab$
\[{(a - b)^3} = {a^3} - {b^3} - 3ab(a - b)\]

Complete step-by-step solution:
Since given that we have two values for the equations $a - b = 3$and ${a^3} - {b^3} = 117$
We solve the two equations with the help of the algebraic formulas to get the required value of $a + b$
Let us start with the first formula which is given as ${a^3} - {b^3} = (a - b)({a^2} + {b^2} + ab)$then substitute the values of $a - b = 3$and ${a^3} - {b^3} = 117$
Thus, we get, ${a^3} - {b^3} = (a - b)({a^2} + {b^2} + ab) \Rightarrow 117 = 3 \times ({a^2} + {b^2} + ab)$
Further solving with the help of division operation, we get $117 = 3 \times ({a^2} + {b^2} + ab) \Rightarrow ({a^2} + {b^2} + ab) = \dfrac{{117}}{3} \Rightarrow 39$
Convert the equation as \[({a^2} + {b^2} + ab) = 39 \Rightarrow {a^2} + {b^2} + ab + ab = 39 + ab\] (adding $ab$ on both sides to get the general equation), thus we get \[{a^2} + {b^2} + ab + ab = 39 + ab \Rightarrow {(a + b)^2} = 39 + ab\]
Now take this \[{(a + b)^2} = 39 + ab\] as the equation $(1)$
Again, from the given that we have, $a - b = 3$and now cubing both sides of the equations we get, \[a - b = 3 \Rightarrow {(a - b)^3} = {3^3}\]
Further solving we get, \[{(a - b)^3} = {3^3} \Rightarrow {a^3} - {b^3} - 3ab(a - b) = 27\] (from the last formula)
As we know the value of ${a^3} - {b^3} = 117$and substituting this we get, \[{a^3} - {b^3} - 3ab(a - b) = 27 \Rightarrow 117 - 3ab(a - b) = 27\]and also, we know that $a - b = 3$
Thus, we get, \[117 - 3ab(a - b) = 27 \Rightarrow 117 - 3ab(3) = 27\]
Solving the equations step by step, we get \[117 - 3ab(3) = 27 \Rightarrow - 9ab = - 90 \Rightarrow ab = 10\]
Thus, we get the value of $ab = 10$and substituting the value in the equation $(1)$then we get \[{(a + b)^2} = 39 + ab \Rightarrow {(a + b)^2} = 39 + 10 \Rightarrow {(a + b)^2} = 49\]
Applying the square root on both sides we get, \[{(a + b)^2} = 49 \Rightarrow (a + b) = \sqrt {49} \Rightarrow \pm 7\](taking out the values from the square root will get positive or negative)
Therefore, we get, \[(a + b) = \pm 7\]

Note:Since we can show that all formulas used are equals on left and right sides as taking $(a - b)({a^2} + {b^2} + ab)$then solve further, we get $(a - b)({a^2} + {b^2} + ab) = {a^3} - {b^3} + a{b^2} - a{b^2} + {a^2}b - {a^2}b$ then we get after canceling the common terms, \[(a - b)({a^2} + {b^2} + ab) = {a^3} - {b^3}\]and hence the values are equal at left and right sides.
Some general identities are
${(a + b)^2} = {a^2} + {b^2} + 2ab$
${(a + b)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}$