Sorry, but no top university ever asks questions about basic surd manipulation. They might ask about the general principle. They might ask about algebraic manipulation. This is an exercise that is tortuous and maybe good for practice, but pretending it’s university entrance level is misleading. An able 14 year old could be expected to do it, once shown the principle of rationalising the denominator.
Reducing the expression without using the conjugate:
(√3 + √2)(√3 - √2) = 1
(√3 - √2) = 1/ (√3 + √2)
(√3 + √2 + 1) / (√3 - √2 + 1)
= (√3 + √2 + 1) / (1 / [√3 +√2] + 1) // substitute
= (√3 + √2 + 1) (√3 + √2) / (√3 + √2 + 1) // motivates the multiplication
= (√3 + √2)
That's right
What is the natural log of "I"?
Math Interview Tricks: [(√18 + √12 + √6)/(√18 - √12 + √6)]⁷ =?
(√18 + √12 + √6)/(√18 - √12 + √6) = (√3 + √2 + 1)/(√3 - √2 + 1)
= [(√3 + √2 + 1)(√3 + √2 - 1)]/[(√3 - √2 + 1)(√3 + √2 - 1)]
= [(√3 + √2)² - 1]/[3 - (√2 - 1)²] = (4 + 2√6)/(2√2) = √2 + √3
(√2 + √3)² = 5 + 2√6, (√2 + √3)⁴ = [(√2 + √3)²]² = (5 + 2√6)² = 49 + 20√6
(√2 + √3)³ = (√2 + √3)[(√2 + √3)²] = (√2 + √3)(5 + 2√6) = 11√2 + 9√3
(√2 + √3)⁷ = [(√2 + √3)³][(√2 + √3)⁴] = (11√2 + 9√3)(49 + 20√6)
= 539√2 + 441√3 + 440√3 + 540√2 = 1079√2 + 881√3
[(√18 + √12 + √6)/(√18 - √12 + √6)]⁷ = (√2 + √3)⁷ = 1079√2 + 881√3
Another method is multiplying x^4*x^2*x
That's a great shortcut! 👍
Sorry, but no top university ever asks questions about basic surd manipulation. They might ask about the general principle. They might ask about algebraic manipulation.
This is an exercise that is tortuous and maybe good for practice, but pretending it’s university entrance level is misleading. An able 14 year old could be expected to do it, once shown the principle of rationalising the denominator.