We will solve the problems of this exercise by using various Algebraic Formulae. Different formulae used are listed below with examples thereof.

- (a + b)
^{2}+ (a – b)^{2}= 2(a^{2}+ b^{2}) and

(a + b)^{2} – (a – b)^{2} = 4ab

Example:

If a + b = 7 and a – b = 3 then find the value of a^{2} + b^{2} and ab

Now we know that

(a + b)^{2} + (a – b)^{2} = 2(a^{2} + b^{2})

Putting the values of (a + b) and (a – b)

= (7)^{2} + (3)^{2} = 2(a^{2} + b^{2})

= 49 + 9 = 2(a^{2} + b^{2}) = 58

Hence a^{2 }+ b^{2 }= 58/2 = 29

Similarly we know that

(a + b)^{2} – (a – b)^{2} = 4 ab

=(7)^{2} – (3)^{2} = 4 ab

= 49 – 9 = 4 ab = 40

Hence ab = 40/4 = 10

**Answer: a ^{2} + b^{2} = 29 and ab = 10**

- (a + b + c)
^{2}= a^{2}+ b^{2}+ c^{2}+ 2ab +2ac + 2bc

Example:

If a^{2} + b^{2} + c^{2 }= 43 and ab + bc + ca = 3

Find the value of a + b + c

We have the formula

(a + b + c)^{2} = (a^{2} + b^{2} + c^{2}) + 2(ab + ac + bc)

Putting the values given in the example

(a + b+ c)^{2} = (43) + 2(3) = 43 + 6 = 49

Taking the Square Root of both sides

**(a + b + c) = 7 Answer**

- (a +b)
^{3}= a^{3}+ 3ab(a + b) + b^{3}

(a – b)^{3} = a^{3} – 3ab(a – b) – b^{3}

Example:

If 2x – 3y = 10 and xy = 2 then find 8x^{3} – 27y^{3}

We have the folrmula

(a – b)^{3} = a^{3} – 3ab(a – b) – b^{3}

= (2x – 3y)^{3} = (2x)^{3} – 3 . 2x . 3y (10) – (3y)^{3} = (10)^{3}

= 8x^{3} – 18xy (10) – 27y^{3} = 1000 = 8x^{3 }– 18 . 2 . 10 – 27 y^{3}

= 8x^{3} – 27 y^{3} – 360 = 1000

**= 8x ^{3} – 27y^{3} = 1360 Answer**

- a
^{3}+ b^{3}= (a + b) (a^{2}– ab + b^{2})

a^{3} – b^{3} = (a – b) (a^{2} + ab + b^{2})

The questions on these formulae may be solved in the same manner

**QUESTIONS**

1 | 2 | 3 |

4 | 5 | 6 |

7 | 8 | 9 |

10 | 11 | 12 |

13 | 14 | 15 (I) |

15(II) | 15(III) | 15(IV) |