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Show that : a) (9p-5q)square+(180pq=(9p+5q)square
?
Most Upvoted Answer
Show that : a) (9p-5q)square+(180pq=(9p+5q)square Related: Examples: ...
LHS = (9p – 5q)2 + 180pq
=(9p)2 – 2(9p)(5q) + (5q)2 + 180pq
=81p2 – 90pq + 25q2 + 180pq
=81p2 + (–90 + 180)pq + 25q2
=81p2 + 90pq + 25q2
RHS = (9p + 5q)2
=(9p)2 + 2(9p)(5q) + (5q)2
=81p2 + 90pq + 25q2
Since LHS=RHS, Proved
Community Answer
Show that : a) (9p-5q)square+(180pq=(9p+5q)square Related: Examples: ...
To prove that (9p - 5q)^2 = (9p)^2 - 2(9p)(5q) + (5q)^2, let's expand both sides of the equation.

Expanding the left side:

(9p - 5q)^2 = (9p - 5q)(9p - 5q)
= (9p)(9p) - (9p)(5q) - (5q)(9p) + (5q)(5q)
= 81p^2 - 45pq - 45pq + 25q^2
= 81p^2 - 90pq + 25q^2

Expanding the right side:

(9p)^2 - 2(9p)(5q) + (5q)^2 = 81p^2 - 90pq + 25q^2

Since both sides of the equation are equal, we have proved that (9p - 5q)^2 = (9p)^2 - 2(9p)(5q) + (5q)^2.

To prove that 180pq = (9p)^2 - (5q)^2, let's expand both sides of the equation.

Expanding the left side:

180pq = 180pq

Expanding the right side:

(9p)^2 - (5q)^2 = (9p)(9p) - (5q)(5q)
= 81p^2 - 25q^2

Since both sides of the equation are equal, we have proved that 180pq = (9p)^2 - (5q)^2.

Therefore, we have shown that (9p - 5q)^2 = (9p)^2 - 2(9p)(5q) + (5q)^2 and 180pq = (9p)^2 - (5q)^2.
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Show that : a) (9p-5q)square+(180pq=(9p+5q)square Related: Examples: Algebraic Identities(Part 2)?
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