FREE ALGEBRA HELP - SPECIAL PRODUCTS, EXAMPLES, EXERCISE USING FORMULAS

Please study Algebra Formulas before Free Algebra Help if you have not already done so. There we have listed out all the Algebra Formulas that need to be remembered.

Here we present Solved Examples and Exercise problems on application of those Formulas.

Example 1 of Free Algebra Help

Solve the following problems on Free Algebra Help.

If 3x + 4y = 12 and xy = 2, find (i) 9x^{2} + 16y^{2} (ii) 27x^{3} + 64y^{3}

Solution to the Example 1 on Free Algebra Help : (i) Let P = 9x^{2} + 16y^{2} = (3x)^{2} + (4y)^{2} This looks like a^{2} + b^{2} with 3x in place of a and 4y in place of b So the problem becomes finding a^{2} + b^{2} given (a + b) and ab{= (3x)(4y) = 12xy = 12(2)} We know a^{2} + 2ab + b^{2} = (a + b)^{2} (See Formula 1) ⇒ a^{2} + b^{2} = (a + b)^{2} - 2ab Applying this here to find P, we get P = (3x)^{2} + (4y)^{2} = (3x + 4y)^{2} - 2(3x)(4y) = (3x + 4y)^{2} - 24xy By data, 3x + 4y = 12 and xy = 2 Using these values here to find P, we get P = (12)^{2} - 24(2) = 144 - 48 = 96. Ans.

(ii) Let P = 27x^{3} + 64y^{3} we know 27 = 3 x 3 x 3 = 3^{3} ⇒ 27x^{3} = 3^{3}x^{3} = (3x)^{3}; 64 = 4 x 4 x 4 = 4^{3} ⇒ 64y^{3} = 4^{3}y^{3} = (4y)^{3}; ∴ P = (3x)^{3} + (4y)^{3} This looks like a^{3} + b^{3} with 3x in place of a and 4y in place of b So the problem becomes finding a^{3} + b^{3} given (a + b) and ab {= (3x)(4y) = 12xy = 12(2)} We know a^{3} + 3ab(a + b) + b^{3} = (a + b)^{3} (See Formula 6) ⇒ a^{3} + b^{3} = (a + b)^{3} - 3ab(a + b) Applying this here to find P, we get P = (3x)^{3} + (4y)^{3} = (3x + 4y)^{3} - 3(3x)(4y)(3x + 4y) = (3x + 4y)^{3} - 36xy(3x + 4y) By data, 3x + 4y = 12 and xy = 2 Using these values here to find P, we get P = (12)^{3} - 36(2)(12) = 144 x 12 - 144 x 6 = 144(12 - 6) = 144 x 6 = 864. Ans.

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Solve the following problems on Free Algebra Help.

If (x - 1⁄x) = 4, find (i) x^{2} + 1⁄x^{2} (ii) x^{4} + 1⁄x^{4} (iii) x^{3} - 1⁄x^{3}

Solution to the Example 2 on Free Algebra Help : (i) Let P = x^{2} + 1⁄x^{2} = x^{2} + (1⁄x)^{2} This looks like a^{2} + b^{2} with x in place of a and (1⁄x) in place of b So the problem becomes finding a^{2} + b^{2} given (a - b) and ab {= (x)(1⁄x) = 1} We know a^{2} - 2ab + b^{2} = (a - b)^{2} (See Formula 2) ⇒ a^{2} + b^{2} = (a - b)^{2} + 2ab Applying this here to find P, we get P = x^{2} + (1⁄x)^{2} = (x - 1⁄x)^{2} + 2(x)(1⁄x) = (x - 1⁄x)^{2} + 2(1) = (x - 1⁄x)^{2} + 2 By data, (x - 1⁄x) = 4 Using this value here to find P, we get P = (4)^{2} + 2 = 16 + 2 = 18 ∴ x^{2} + 1⁄x^{2} = 18. Ans.

(ii) Let P = x^{4} + 1⁄x^{4} = (x^{2})^{2} + (1⁄x^{2})^{2} This looks like a^{2} + b^{2} with (x^{2}) in place of a and (1⁄x^{2}) in place of b So the problem becomes finding a^{2} + b^{2} given (a + b) { = x^{2} + 1⁄x^{2} = 18 from (i) } and ab {= (x^{2})(1⁄x^{2}) = 1} We know a^{2} + 2ab + b^{2} = (a + b)^{2} (See Formula 1) ⇒ a^{2} + b^{2} = (a + b)^{2} - 2ab Applying this here to find P, we get P = (x^{2})^{2} + (1⁄x^{2})^{2} = (x^{2} + 1⁄x^{2})^{2} - 2(x^{2})(1⁄x^{2}) = (x^{2} + 1⁄x^{2})^{2} - 2(1) = (x^{2} + 1⁄x^{2})^{2} - 2 From the result found in (i), (x^{2} + 1⁄x^{2}) = 18. Using this value here to find P, we get P = (18)^{2} - 2 = 324 - 2 = 322 ∴ x^{4} + 1⁄x^{4} = 322. Ans.

(iii) Let P = x^{3} - 1⁄x^{3} = x^{3} - (1⁄x)^{3} This looks like a^{3} - b^{3} with x in place of a and (1⁄x) in place of b So the problem becomes finding a^{3} - b^{3} given (a - b) { = (x - 1⁄x) = 4 by data} and ab {= (x)(1⁄x) = 1} We know a^{3} - 3ab(a - b) - b^{3} = (a - b)^{3} (See Formula 7) ⇒ a^{3} - b^{3} = (a - b)^{3} + 3ab(a - b)Applying this here to find P, we get P = (x)^{3} - (1⁄x)^{3} = (x - 1⁄x)^{3} + 3(x)(1⁄x)(x - 1⁄x) = (x - 1⁄x)^{3} + 3(1)(x - 1⁄x) By data, x - 1⁄x= 4 Using these values here to find P, we get P = (4)^{3} + 3(4) = 64 + 12 = 76. Ans.

Solve the following problems on Free Algebra Help.

Find (i) 997^{2} (ii) 997^{3} using Algebra Formulas

Solution to the Example 3 on Free Algebra Help : (i) Let P = 997^{2} 997 is nearer to 1000. We can write 997 = 1000 - 3. ∴ P = (1000 - 3)^{2}; This looks like (a - b)^{2} with (1000) in place of a and (3) in place of b We have (a - b)^{2} = a^{2} - 2ab + b^{2} (See Formula 2) ∴ P = (1000 - 3)^{2} = (1000)^{2} - 2(1000)(3) + (3)^{2}= 1000000 - 6000 + 9 = 1000(1000 - 6) + 9 = 1000(994) + 9 = 994000 + 9 = 994009. Ans.

(ii) Let P = 997^{3} = (1000 - 3)^{3} This looks like (a - b)^{3} with (1000) in place of a and (3) in place of b We have (a - b)^{3} = a^{3} - 3a^{2}b + 3ab^{2} - b^{3} (See Formula 7) ∴ P = (1000 - 3)^{3} = (1000)^{3} - 3(1000)^{2}(3) + 3(1000)(3)^{2} - (3)^{3} = (1000) {(1000)^{2} - 9000 + 27} - 27 = (1000) [(1000){(1000) - 9} + 27] - 27 = (1000) [(1000){991} + 27] - 27 = (1000) [{991000} + 27] - 27 = (1000) [{991027}] - 27 = 991027000 - 27 = 991026973. Ans.

Solve the following problems on Free Algebra Help.

If 2x - 3y = 10 and xy = 3, find (i) 4x^{2} + 9y^{2} (ii) 8x^{3} - 27y^{3}

If x + 1⁄x = 5, find (i)x^{2} + 1⁄x^{2} (ii) x^{4} + 1⁄x^{4} (iii) x^{3} + 1⁄x^{3}

Find (i) 1002^{2} (ii) 1002^{3} using Algebra Formulas

For Answers See at the bottom of the page.

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