Solve left(1999-xright)^3+left(x-1998right)^3=1

$${ \left(1999-x \right) }^{ 3 } + { \left(x-1998 \right) }^{ 3 } =1$$

Solve for x

$$x=1998$$

$$x=1999$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(1999-x\right)^{3}.

$$7988005999-11988003x+5997x^{2}-x^{3}+\left(x-1998\right)^{3}=1$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-1998\right)^{3}.

$$7988005999-11988003x+5997x^{2}-x^{3}+x^{3}-5994x^{2}[phone]x-7976023992=1$$

Combine -x^{3} and x^{3} to get 0.

$$7988005999-11988003x+5997x^{2}-5994x^{2}[phone]x-7976023992=1$$

Combine 5997x^{2} and -5994x^{2} to get 3x^{2}.

$$7988005999-11988003x+3x^{2}[phone]x-7976023992=1$$

Combine -11988003x and 11976012x to get -11991x.

$$7988005999-11991x+3x^{2}-7976023992=1$$

Subtract 7976023992 from 7988005999 to get 11982007.

$$11982007-11991x+3x^{2}=1$$

Subtract 1 from both sides.

$$11982007-11991x+3x^{2}-1=0$$

Subtract 1 from 11982007 to get 11982006.

$$11982006-11991x+3x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$3x^{2}-11991x+11982006=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -11991 for b, and 11982006 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11991\right)±\sqrt{\left(-11991\right)^{2}-4\times 3\times 11982006}}{2\times 3}$$

Square -11991.

$$x=\frac{-\left(-11991\right)±\sqrt{143784081-4\times 3\times 11982006}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-11991\right)±\sqrt{143784081-12\times 11982006}}{2\times 3}$$

Multiply -12 times 11982006.

$$x=\frac{-\left(-11991\right)±\sqrt{143784081-143784072}}{2\times 3}$$

Add 143784081 to -143784072.

$$x=\frac{-\left(-11991\right)±\sqrt{9}}{2\times 3}$$

Take the square root of 9.

$$x=\frac{-\left(-11991\right)±3}{2\times 3}$$

The opposite of -11991 is 11991.

$$x=\frac{11991±3}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{11991±3}{6}$$

Now solve the equation x=\frac{11991±3}{6} when ± is plus. Add 11991 to 3.

$$x=\frac{11994}{6}$$

Divide 11994 by 6.

$$x=1999$$

Now solve the equation x=\frac{11991±3}{6} when ± is minus. Subtract 3 from 11991.

$$x=\frac{11988}{6}$$

Divide 11988 by 6.

$$x=1998$$

The equation is now solved.

$$x=1999 x=1998$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(1999-x\right)^{3}.

$$7988005999-11988003x+5997x^{2}-x^{3}+\left(x-1998\right)^{3}=1$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-1998\right)^{3}.

$$7988005999-11988003x+5997x^{2}-x^{3}+x^{3}-5994x^{2}[phone]x-7976023992=1$$

Combine -x^{3} and x^{3} to get 0.

$$7988005999-11988003x+5997x^{2}-5994x^{2}[phone]x-7976023992=1$$

Combine 5997x^{2} and -5994x^{2} to get 3x^{2}.

$$7988005999-11988003x+3x^{2}[phone]x-7976023992=1$$

Combine -11988003x and 11976012x to get -11991x.

$$7988005999-11991x+3x^{2}-7976023992=1$$

Subtract 7976023992 from 7988005999 to get 11982007.

$$11982007-11991x+3x^{2}=1$$

Subtract 11982007 from both sides.

$$-11991x+3x^{2}=1-11982007$$

Subtract 11982007 from 1 to get -11982006.

$$-11991x+3x^{2}=-11982006$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$3x^{2}-11991x=-11982006$$

Divide both sides by 3.

$$\frac{3x^{2}-11991x}{3}=-\frac{11982006}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{11991}{3}\right)x=-\frac{11982006}{3}$$

Divide -11991 by 3.

$$x^{2}-3997x=-\frac{11982006}{3}$$

Divide -11982006 by 3.

$$x^{2}-3997x=-3994002$$

Divide -3997, the coefficient of the x term, by 2 to get -\frac{3997}{2}. Then add the square of -\frac{3997}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3997x+\left(-\frac{3997}{2}\right)^{2}=-3994002+\left(-\frac{3997}{2}\right)^{2}$$

Square -\frac{3997}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3997x+\frac{15976009}{4}=-3994002+\frac{15976009}{4}$$

Add -3994002 to \frac{15976009}{4}.

$$x^{2}-3997x+\frac{15976009}{4}=\frac{1}{4}$$

Factor x^{2}-3997x+\frac{15976009}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3997}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3997}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{3997}{2}=\frac{1}{2} x-\frac{3997}{2}=-\frac{1}{2}$$

Add \frac{3997}{2} to both sides of the equation.

$$x=1999 x=1998$$