Free sample · Sec 3–4 A-Math

Algebraic identities

Given x+1/x=3

x + \frac{1}{x} = 3

, find x^3+ 1x^3

x^{3} + \frac{1}{x ^{3}}

. I Do, then We Do, then You Do.

Primary volume · P5 · Cost · Sec 1 · Algebra · Sec 1 · Logic · Sec 1 · Angles

I Do

Watch the method

Given x+1/x=3 $x + \frac{1}{x} = 3$ . Find x^3+ 1x^3 $x^{3} + \frac{1}{x ^{3}}$ .

Square first: (x+1/x)^2=x^2+2+ 1x^2 $(x + \frac{1}{x})^{2} = x^{2} + 2 + \frac{1}{x ^{2}}$ , so 9=x^2+ 1x^2+2 $9 = x^{2} + \frac{1}{x ^{2}} + 2$ and x^2+ 1x^2=7 $x^{2} + \frac{1}{x ^{2}} = 7$ . Why square? It clears the linear sum into the quadratic sum you need next.

Cube identity: (x+1/x)^3=x^3+ 1x^3+3(x+1/x) $(x + \frac{1}{x})^{3} = x^{3} + \frac{1}{x ^{3}} + 3 (x + \frac{1}{x})$ . Plug in 3 $3$ : 27=x^3+ 1x^3+9 $27 = x^{3} + \frac{1}{x ^{3}} + 9$ , so x^3+ 1x^3=18 $x^{3} + \frac{1}{x ^{3}} = 18$ .

We Do

1.Given x+1/x=3 $x + \frac{1}{x} = 3$ . What is x^2+ 1x^2 $x^{2} + \frac{1}{x ^{2}}$ ?

You Do

Do it alone

Same given x+1/x=3 $x + \frac{1}{x} = 3$ . Exact value of x^3+ 1x^3 $x^{3} + \frac{1}{x ^{3}}$ ?

The full chapter continues from here

Partial fractions, modulus, binomial, and exam papers unlock after the parent PayNow subscription.

Enrol · Pricing